Answer:
B. Cl -1
Explanation:
Chlorine has 7 valence electrons and needs one more to complete its octet thus takes one electron from sodium to become negatively charged Cl^-1.
Acetone is a solvent used in fingernail polish remover. Acetone is polar. It will form
solutions, or be soluble, in all the following except
O A. ethanol, which has molecules that have
one end with a slight charge
B. ammonia, which is made of molecules with
a slightly negative nitrogen atom and three
slightly positive hydrogen atoms
O C. water, which has oxygen atoms with a
slight negative charge
D. vegetable oil, which has atoms that share
electrons equally, so that there is no charge
difference
Read Question
Correct:
Answer:
d
Explanation:
Olive oil has a density of 0.92g/ml. How much would 1.0 Liter of olive oil weigh in grams?
Answer:
9.2x10²g
Explanation:
Data obtained from the question include the following:
Density = 0.92g/ml
Volume = 1L = 1 x 1000 = 1000mL
Mass =..?
Density is simply defined as the mass of the substance per unit volume of the substance. Mathematically it can be represented as:
Density = Mass /volume.
Mass = Density x volume
Mass = 0.92 x 1000
Mass = 9.2x10²g.
Therefore, 1L of olive will weigh 9.2x10²g.
What is the kinetic energy,in J,of an Ar atom moving at a speed of 650 m/s
Answer:
1.40 × 10⁻²⁰ J
Explanation:
Step 1: Calculate the mass of 1 atom of argon
The molar mass of argon is 39.95 g/mol, that is, 6.02 × 10²³ atoms of Ar have a mass of 39.95 g. We can use this relation to find the mass of 1 atom of Ar.
[tex]\frac{39.95g}{1mol} \times \frac{1mol}{6.02 \times 10^{23}atom } =6.64 \times 10^{-23}g/atom[/tex]
Step 2: Convert the mass of 1 atom of argon to kilograms
We will use the relationship 1 kg = 1,000 g.
[tex]6.64 \times 10^{-23}g \times \frac{1kg}{1,000g} =6.64 \times 10^{-26}kg[/tex]
Step 3: Calculate the kinetic energy of 1 atom of Ar moving at 650 m/s
[tex]E=\frac{1}{2} \times m \times v^{2} = \frac{1}{2} \times 6.64 \times 10^{-26}kg \times (650m/s)^{2} = 1.40 \times 10^{-20}J[/tex]
The kinetic energy of an Ar atom at a speed of 650 m/s is 1.40 × 10⁻²⁰ J
Kinetic Energy:
Kinetic energy is directly proportional to the mass of the object and to the square of its velocity:
K.E = [tex]\frac{1}{2}[/tex] x [tex]m[/tex] x [tex]v^{2}[/tex]
Step 1: Calculate the mass of 1 atom of argon
The molar mass of argon is 39.95 g/mol, that is, 6.02 × [tex]10^{-23}[/tex] atoms of Ar have a mass of 39.95 g. We can use this relation to find the mass of 1 atom of Ar.
[tex]\frac{39.95 g}{1 mol} * \frac{1 mol}{1,000 g} = 6.64 * 10^{-26} g/ atom[/tex]
Step 2: Convert the mass of 1 atom of argon to kilograms
We will use the relationship 1 kg = 1,000 g.
6.64 x [tex]10^{-23}[/tex] g x [tex]\frac{1 kg}{1,000g}[/tex] = 6.64 x [tex]10^{-26}[/tex] kg
Step 3: Calculate the kinetic energy of 1 atom of Ar moving at 650 m/s
E= [tex]\frac{1}{2}[/tex] x [tex]m[/tex] x [tex]v^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 6.64 x [tex]10^{-26}[/tex] kg x (650 m/s)² = 1.40 x [tex]10^{-20}[/tex] J
Learn more:
https://brainly.com/question/12337396
What is the color of litmus solution in sodium hydroxide?
Answer:
Sodium Hydroxide turns blue litmus red .
Based on the activity series provided which reactants will form products? F>Cl>Br>l
Answer:
CuI₂ + Br₂
Explanation:
Answer:
a
Explanation:
A boy reached his home by riding motorcycle within 5 minutes by travelling 8 km distance ,them what is his average velocity?
Answer:
1.6
Explanation:
velocity=distance/time
=8/5
The pressure exerted by 1.5 mol of gas in a 13 L flask at 22 °C is ____ kPa
Answer:
282.7KPa
Explanation:
Step 1:
Data obtained from the question.
Number of mole of (n) = 1.5 mole
Volume (V) = 13L
Temperature (T) = 22°C = 22 + 273°C = 295K
Pressure (P) =..?
Gas constant (R) = 0.082atm.L/Kmol
Step 2:
Determination of the pressure exerted by the gas.
This can be obtained by using the ideal gas equation as follow:
PV = nRT
P = nRT /V
P = 1.5 x 0.082 x 295 / 13
P = 2.79atm.
Step 3:
Conversion of 2.79atm to KPa.
This is illustrated below:
1 atm = 101.325KPa
Therefore, 2.79atm = 2.79 x 101.325 = 282.7KPa
Therefore, the pressure exerted by the gas in KPa is 282.7KPa
Pressure is defined as the force exerted by the molecules colliding with the surface of the object and other surfaces. The ideal gas follows the general gas law equation, PV = nRT.
The pressure exerted by the gas is 282.7 KPa.
Given:
Pressure (P) = ?
Gas constant (R) = 0.082atm.L/Kmol
Temperature (T) = 22°C = 22 + 273°C = 295K
Volume (V) = 13L
Number of mole of (n) = 1.5 mole
Now, using the ideal gas equation:
PV = nRT
Substituting the values, we get:
P x 13 = 1.5 x 0.082 x 295
P x 13 = 36.285
P = [tex]\dfrac{36.285}{13}[/tex]
Pressure = 2.79 atm.
Also, 1 atm = 101.32 KPa
2.79 atm = 2.79 x 101.32 = 282.7 KPa.
Thus, the pressure exerted by the gas molecules is 282.7 KPa.
To know more about ideal gas law, refer to the following link:
https://brainly.com/question/12124605
A galvanic (voltaic) cell consists of an electrode composed of magnesium in a 1.0 M magnesium ion solution and another electrode composed of silver in a 1.0 M silver ion solution, connected by a salt bridge. Calculate the standard potential for this cell at 25 °C.
Answer:
3.17 V
Explanation:
The cell is operating under standard conditions. These standard conditions include; that the reaction takes place at 298 Kelvin (room temperature), the pressure of the system is 1 atmosphere (standard pressure), and the solutions have a Molarity of 1.0 M for both the anode and cathode solutions. All these conditions are satisfied in the cell under review in the question.
Hence;
E°anode (magnesium)= -2.37 V
E°cathode (silver) = 0.80 V
E°cell= E°cathode -E°anode
E°cell= 0.80-(-2.37)
E°cell= 0.80 + 2.37
E°cell= 3.17 V
Hence the standard cell potential of this cell at 25°C is 3.17 V
Answer:
The standard potential at 25ºC is 3.17 V.
Explanation:
The anode in a galvanic cell is the electrode at which oxidation occurs and the cathode is the electrode at which reduction occurs.
The overall cell reaction will be the sum of two half-cell reactions. The standard reduction potentials are:
Mg²⁺ (1.0 M) + 2e⁻ → Mg (s) Eº = -2.37
Ag⁺ (1.0 M) + e⁻ → Ag (s) Eº= +0.80
Since the reactants are in their standard states (1.0 M) and at 25ºC we can write the half-cell reactions as follows:
Anode (oxidation): Mg (s) → Mg²⁺ (1.0 M) + 2e⁻
Cathode (reduction): 2Ag⁺ (1.0 M) + 2e⁻ → 2Ag (s)
Overall: Mg (s) +2 Ag⁺ (1.0 M) + 2e⁻ → 2Ag (s) + Mg²⁺ (1.0 M) + 2e⁻
In order to balance the overall equation we multiply the reduction of Ag⁺ by 2. We can do so because, as an intensive property, E° is not affected by this procedure.
The standard emf of the cell, E°cell , which is composed of a contribution from the anode and a contribution from the cathode, is given by:
[tex] Eº cell = Eº cathode - Eº anode [/tex]
[tex] Eº cell = EºAg⁺/Ag - Eº Mg²⁺/Mg [/tex]
[tex] Eº cell =0.80 V - (-2.37 V) [/tex]
Eº cell = 3.17 V
Combustion analysis of a 13.42-g sample of an unknown organic compound (which contains only carbon, hydrogen, and oxygen) produced 36.86 g CO2 and 10.06 g H2O. The molar mass of the compound is 288.38 g/mol . Part A Find the molecular formula of the unknown compound. Express your answer as a chemical formula.
Answer:
[tex]C_{18}H_{24}O_3[/tex]
Explanation:
Hello,
In this case, combustion analyses help us to determine the empirical formula of a compound via the quantification of the released carbon dioxide and water since the law of conservation of mass is leveraged to attain it. In such a way, as 36.86 g of carbon dioxide were obtained, this directly represents the mass of carbon present in the sample, thus, we first compute the moles of carbon:
[tex]n_{C}=36.86gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molC}{1molCO_2} =0.838molC[/tex]
Then, into the water one could find the moles of hydrogen:
[tex]n_{H_2O}=10.06gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O} =1.12molH[/tex]
Now, we compute the moles of oxygen by firstly computing the mass of oxygen:
[tex]m_{O}=13.42g-36.86gCO_2*\frac{12gC}{44gCO_2} -10.06 gH_2O*\frac{2gH}{18gH_2O} =2.25gO[/tex]
[tex]n_O=2.25gO*\frac{1molO}{16gO} =0.141molO[/tex]
Then, we have the mole ratio:
[tex]C_{0.838}H_{1.12}O_{0.141}\rightarrow C_6H_8O[/tex]
Whose molar mass is 12x6+1x8+16=96 g/mol, but the whole compound molar mass is 288.38, the factor is 288.38/96 =3, Therefore the formula:
[tex]C_{18}H_{24}O_3[/tex]
Regards.
Answer:
Molecular mass = [tex]C_{6} H_{8} O_{16}[/tex]
Explanation:
[tex](\frac{36.86gCO_{2} }{44g/molCO_{2} })[/tex] [tex](\frac{10.06gH_{2}O}{18g/molH_{2}O} )[/tex]
(0.8377mol of CO2) (0.556 mol of H2O)
mole ratio of CO2 to H2O = 1.5 : 1 ≅ 3 : 2
[tex](CO_{2} )_{3} , (H_{2} O)_{2}[/tex]
⇒ [tex](C_{3} H_{4} O_{8} )_{x}[/tex] = 288.38
(12 × 3 + 1 × 4 + 16 × 8)x = 288.38
168x = 288.38
x = 1.71 ≅ 2
∴ Molecular mass = [tex]C_{6} H_{8} O_{16}[/tex]
If the specific heat of gold is 0.13 J/gC, what is the amount of energy (heat) required to raise 30.0g of gold from 15 degrees Celsius to 41 degrees Celsius?
Answer: The amount of energy required is 101 Joules
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=m\times c\times \Delta T[/tex]
Q = Heat absorbed = ?
m= mass of gold = 30.0 g
c = specific heat capacity of gold = [tex]0.13J/g^0C[/tex]
Initial temperature of the water = [tex]T_i[/tex] = 15°C
Final temperature of the water = [tex]T_f[/tex] = 41°C
Change in temperature ,[tex]\Delta T=T_f-T_i=(41-15)^0C=26^0C[/tex]
Putting in the values, we get:
[tex]Q=30.0g\times 0.13J/g^0C\times 26^0C[/tex]
[tex]Q=101J[/tex]
The amount of energy required is 101 Joules
Determine the number of moles of air present in 1.35 L at 750 torr and 17.0°C.
Which equation should you use?
Answer:
0.0560 mol
Explanation:
You need to use the ideal gas law equation:
[tex]PV = nRT[/tex]
P = 750 torr
V = 1.35 L
n = moles
R = 62.364 L torr [tex]mol^{-1}[/tex] [tex]K^{-1}[/tex]
T = 17.0 ˚C + 273.15 = 290.15 K
Rearranging the equation for n:
[tex]n= \frac{PV}{RT}[/tex]
[tex]n= \frac{(750)(1.35)}{(62.364)(290.15)}[/tex]
n = 0.0560 mol
Answer:
n=PV/RT
Explanation:
on edge 2020
4. Alex is a soil specialist and he sells manure best suited for wheat. He goes to a village where farmers are
planning to sow wheat which grows best in clayey and loamy soil. He tests the soil sample and finds
that the soil is more suitable for the growth of cotton rather than wheat. He tells the villagers about
this despite knowing that they will not buy his manure.
(a) Why is clayey-loamy soil preferred for growing wheat?
(b) Which value is shown by Alex?
Please answer the question fast
Answer:
Explanation:
1) Clayey-loamy soil tends to be excellent at retaining water which is critical for proper growth of wheat thus this kind of soil is preferred.
2) Alex is showing honesty as he informing the villagers about the actual truth and reality (he is not basing his answers on judgements rather he is basing it on scientific knowledge and being honest about it)
The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanisms for this reaction are: Mechanism A (1) H2(g)+I2(g)−→k12HI(g)(one-step reaction) Mechanism B (1) I2(g)⥫⥬=k−1k12I(g)(fast, equilibrium) (2) H2(g)+2I(g)−→k22HI(g) (slow) Mechanism C (1) I2(g)⥫⥬=k−1k12I(g)(fast, equilibrium) (2) I(g)+H2(g)−→k2HI(g)+H(g) (slow) (3) H(g)+I(g)−→k3HI(g) (fast) Which of these mechanisms are consistent with the observed rate law? mechanism A mechanism C mechanism B In 1967, J. H. Sullivan showed that this reaction was dramatically catalyzed by light when the energy of the light was sufficient to break the I−I bond in an I2 molecule. Which mechanism or mechanisms are consistent with both the rate law and this additional observation? mechanism A mechanism C mechanism B
Answer:
Mechanism A and B are consistent with observed rate law
Mechanism A is consistent with the observation of J. H. Sullivan
Explanation:
In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.
In the proposed mechanisms:
Mechanism A
(1) H2(g)+I2(g)→2HI(g)(one-step reaction)
Mechanism B
(1) I2(g)⇄2I(g)(fast, equilibrium)
(2) H2(g)+2I(g)→2HI(g) (slow)
Mechanism C
(1) I2(g) ⇄ 2I(g)(fast, equilibrium)
(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)
(3) H(g)+I(g)→HI(g) (fast)
The rate laws are:
A: rate = k₁ [H2] [I2]
B: rate = k₂ [H2] [I]²
As:
K-1 [I]² = K1 [I2]:
rate = k' [H2] [I2]
Where K' = K1 * K2
C: rate = k₁ [H2] [I]
As:
K-1 [I]² = K1 [I2]:
rate = k' [H2] [I2]^1/2
Thus, just mechanism A and B are consistent with observed rate law
In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just mechanism A is consistent with the observation of J. H. Sullivan
he Punnett square shows the possible genotype combinations for the offspring of two parents. What are the genotypes of the parents? Gg and Gg GG and GG gg and gg Gg and gg
Answer:
Its A! Gg and Gg
Explanation:
Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s) How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?
Answer:
The correct answer is is option B
b. 93.3 g
Explanation:
SEE COMPLETE QUESTION BELOW
Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s)
How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?
a. 7.30 g
b. 93.3 g
c. 146 g
d. 150 g
e. 196 g
CHECK THE ATTACHMENT FOR STEP BY STEP EXPLANATION
Write the identity of the missing nucleus for the following nuclear decay reaction:
232 Th 228Ra+?
90
88
low.
Express your answer as a particle.
View Available Hints
Hint 1. Determine the missing mass number
1
Hint 2. Determine the missing atomic number
Hint 3. Determine the symbol of the new nucleus
?
Answer:
Alpha decay: 4 2He
Explanation:
Considering the equation given in the question, we can see that the mass number of daughter nuclei (228 88Ra) reduce by 4 and the atomic number reduce by 2 when compared with the parent nucleus (232 90Th). This simply indicates that the parent nucleus is undergoing alpha decay.
Therefore, the missing nucleus is
4 2He i.e alpha decay.
Please see attachment photo for further details.
Where does the stored energy in these cabbage leaves come from ?
Answer:
The energy in cabbage leaves comes from the light from the sun.
Explanation:
The plant keeps the light (in the leaves) as energy to help make it grow.
In the laboratory hydrogen gas is usually made by the following reaction:Zn(s)+2HCl(aq)=H2(g)+ZnCl2(aq)How many liters of H2 gas , collected over water at an atmospheric pressure of 752 mmHg and a temperture of 21 Co, can made from 3.566 g of Zn and excess HCl?the partial pressure of water vapor is 18.65mmHg at 21C0.A) 0.68LB) 2.72LC) 1.36LD) 1.33LE) 0.0975L
Answer:
The correct answer is option C, that is, 1.36 L.
Explanation:
The reaction mentioned in the question is:
Zn (s) + 2HCl (aq) ⇒ H2 (g) + ZnCl2 (aq)
It is clear that one mole of zinc is generating one mole of hydrogen gas as seen in the reaction. The mass of zinc mentioned in the question is 3.566 grams, the no of moles can be determined by using the formula,
n = mass / molecular mass
The molecular mass of zinc is 65.39 g/mol, now putting the values in the formula we get,
n = 3.566 g/ 65.39 g/mol
= 0.0545 mol
Based on the question, the partial pressure of water vapor is 18.65 mmHg and the atmospheric pressure is 752 mmHg. Therefore, the pressure of hydrogen gas will be,
Pressure of hydrogen gas (H2) = 752 mmHg - 18.65 mmHg
= 733.35 mmHg
The liters of hydrogen gas produced can be calculated by using the equation, PV =nRT
R is the gas constant, having the value 62.36 L mmHg/K/mol and T is the temperature (273 + 21 = 294 K).
Now putting the values in the equation we get,
733.35 mmHg * V = 0.0545 mol * 62.36 mmHg/K/mol * 294 K
= 999.19 L / 733.35
= 1.36 L
Hence, the volume of hydrogen gas is 1.36 L.
Structurwl formula for (CH3)2CHCH2OH
Answer:
Isobutanol (IUPAC nomenclature: 2-methylpropan-1-ol)
Explanation:
Isobutanol (IUPAC nomenclature: 2-methylpropan-1-ol) is an organic compound with the formula (CH3)2CHCH2OH (sometimes represented as i-BuOH). Hope this helped!
Assign priorities in the following set of substituents according to Cahn-Ingold-Prelog rules.-Cl, -OH, -CH_2OH, -CH_2SHA B C D(Provide your ranking through a string like abed, starting with the one with the highest priority. Your answer does not need to be capitalized.)
Answer:
1) [tex]-Cl[/tex]
2) [tex]-OH[/tex]
3) [tex]-CH_2SH[/tex]
4) [tex]-CH_2OH[/tex]
Explanation:
We have the substituents:
a) [tex]-Cl[/tex]
b) [tex]-OH[/tex]
c) [tex]-CH_2OH[/tex]
d) [tex]-CH_2SH[/tex]
If we remember that Cahn-Ingold-Prelog rules the highest priority is given by the atomic number. Therefore the highest priority is "Cl" (an atomic number equal to 17), the next one is "OH" due to the oxygen (an atomic number equal to 8). For c) and d) we have a carbon bonded to the chiral carbon, therefore we have to check the next atom. The difference between c) and d) are the "O" and "S" atoms, the atom with the highest atomic number is "S" (an atomic number equal to 16) therefore the highest priority is for d) and then c). So finally, the priority is:
1) [tex]-Cl[/tex]
2) [tex]-OH[/tex]
3) [tex]-CH_2SH[/tex]
4) [tex]-CH_2OH[/tex]
I hope it hepls!
Which hormones are secreted from the posterior pituitary gland?
a. Growth hormone
b. Oxytocin
C. ACTH
d. PTH
the answer is b
The posterior pituitary secretes two important endocrine hormones—oxytocin and antidiuretic hormone
When of benzamide are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing point of pure . On the other hand, when of iron(III) chloride are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for iron(III) chloride in . Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.
The given question is incomplete, the complete question is:
When 238.g of benzamide (C7H7NO) are dissolved in 600.g of a certain mystery liquid X, the freezing point of the solution is 5.0°C lower than the freezing point of pure X. On the other hand, when 238.g of iron(III) chloride are dissolved in the same mass of X, the freezing point of the solution is 11.5°C lower than the freezing point of pure X.
Calculate the van't Hoff factor for iron(III) chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits.
Answer:
The correct answer is 3.0.
Explanation:
Based on the given question, the weight of benzamide given is 238 grams, the molecular mass of benzamide is 121.14 gram per mole. The benzamide is dissolved in 600 grams of liquid X, therefore, the solution's total weight is,
= 238 + 600 = 838 grams
In case of benzamide, ΔTf = kf × molality (It is given that the solution's freezing point is 5 degree C lesser in comparison to the pure X's freezing point). Now putting the values we get,
5 = kf × 238 / (121.14×838) × 1 (Van't Hoff factor) -------- (i)
In case of benzamide, the van't Hoff factor will be 1, as it is neither associate nor dissociate.
On the other hand, the mass of ferric chloride given is 238 grams getting dissolved in the similar mass of X, therefore again the total mass of the solution will be 838 grams. The freezing point of the solution is 11.5 degree C lesser than the pure X's freezing point. The molecular mass of ferric chloride is 162.2 gram per mol.
For FeCl3,
ΔTf = kf × molality
11.5 = kf × 238/ (162.2 × 838) × i (Van't Hoff factor)------ (ii)
Now dividing equation (ii) by (i) we get,
11.5 / 5 = (kf/kf) × (121.14 / 162.2) × i
i = 3.0
A titanium cube contains 3.10•10^23 atoms. The density of a titanium is 4.50g/cm^3. What is the edge length of the cube?
PLEASE HELPPP! :(
Answer:
1.76cm
Explanation:
We'll begin by calculating the mass of titanium that contain 3.10x10²³ atoms. This can be obtained as follow:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10²³ atoms. This implies that 1 mole of titanium also contains 6.02x10²³ atoms.
1 mole of titanium = 48g
Now, if 48g of titanium contains 6.02x10²³ atoms,
Then Xg of titanium will contain 3.10x10²³ atoms i.e
Xg of titanium = (48x3.10x10²³)/6.02x10²³
Xg of titanium = 24.72g
Next, we shall determine the volume of the titanium. This is illustrated below:
Density of titanium = 4.50g/cm³
Mass of titanium = 24.72g
Volume of titanium =..?
Density = Mass /volume
Volume = Mass /Density
Volume of titanium = 24.72g/4.50g/cm³
Volume of titanium = 5.49cm³
Finally we shall determine the edge length of the titanium cube as follow:
Volume = L³
5.49cm³ = L³
Take the cube root of both side
L = 3√(5.49cm³)
L = 1.76cm
Therefore, the edge length is 1.76cm
The pressure in car tires is often measured in pounds per square inch (lb/in.2), with the recommended pressure being in the range of 25 to 45 lb/in.2. Suppose a tire has a pressure of 34.5 lb/in.2 . Convert 34.5 lb/in.2 to its equivalent in atmosphere
Answer:
2.35 atm
Explanation:
1 atm = 14.70 lb/in²
[tex]p = \text{34.5 lb/in}^{2} \times \dfrac{\text{1 atm}}{\text{14.70 lb/in}^{2}} = \textbf{2.35 atm}[/tex]
The potential in an electrochemical cell, E, is related to the Gibb's free energy change (ΔG) for the overall cell redox reaction: (1) ΔG0 = - n F E0 where n is the number of electrons transferred during the redox reaction, F is Faraday's constant (96,500 C / mol), and the superscript 0 indicates standard conditions (1 atm, 1 M concentrations, and 25 °C). Thus, a measurement of the cell voltage at standard conditions can be used to determine ΔG0. As an example, the following cell reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m) generates a cell voltage of +1.10 V under standard conditions. Calculate and enter delta G degree (with 3 sig figs) for this reaction in kJ/mol.
Answer:
Explanation:
As an example, the following cell reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m) generates a cell voltage of +1.10 V under standard conditions. Calculate and enter delta G degree (with 3 sig figs) for this reaction in kJ/mol.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m)
ΔG = ΔG° + RTInQ
Q = 1
ΔG = ΔG°
ΔG = =nFE°
n=no of electrons transfered.
E° = 1.1v
ΔG° = -2 * 96500 * 1.10
= -212300J
ΔG° =-212.3kJ/mol
Therefore, the ΔG° = -212.3kJ/molA flexible container at an initial volume of 4.11 L contains 6.51 mol of gas. More gas is then added to the container until it reaches a final volume of 11.3 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.
Answer:
The added mole will be "11.388 moles".
Explanation:
The given values are:
Initial volume,
V₁ = 4.11 L
Final volume,
V₂ = 11.3 L
Number of moles,
n₁ = 6.51
On applying Avogadro's law,
⇒ [tex]\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}[/tex]
On putting the estimated values, we get
⇒ [tex]\frac{4.11}{6.51} =\frac{11.3}{n_{2}}[/tex]
On applying cross-multiplication, we get
⇒ [tex]4.11 \ n_{2}=11.3\times 6.51[/tex]
⇒ [tex]4.11 \ n_{2}=73.563[/tex]
⇒ [tex]n_{2}=\frac{73.563}{4.11}[/tex]
⇒ [tex]n_{2}=17.898 \ moles[/tex]
So that the number of moles at added gas will be:
[tex]=17.898-6.51[/tex]
[tex]=11.388 \ moles[/tex]
Determine the mass of CaBr2 needed to create a 500. ml solution with a concentration of 1.15 M.
Answer:
[tex]m_{CaBr_2}=115gCaBr_2[/tex]
Explanation:
Hello,
In this case, we remember molarity is defined as the ratio of the moles of the solute to the volume of the solution in litres, therefore, for the given concentration and volume, we should solve the moles of calcium bromide to subsequently compute the mass, as shown below:
[tex]M=\frac{n}{V}\\ \\n=M*V=1.15\frac{mol}{L} *500mL*\frac{1L}{1000mL}=0.575mol[/tex]
Next, since molar mass of calcium bromide is 200 g/mol, we can compute the mass:
[tex]m_{CaBr_2}=0.575mol*\frac{1mol}{200g} \\\\m_{CaBr_2}=115gCaBr_2[/tex]
Regards.
After doing a correct and careful recrystallization, you isolate your product by filtration. Which of the following will be present in the mother liquor? (select all that apply)
a. Impurities
b. Large amount of product
c. Recrystallizing solvent
d. Small amount of produc
Answer:
a. Impurities
c. Recrystallizing solvent
Explanation:
In this type of reaction the products are never considered totally pure, that is why as a final product it must always be taken into account that it is proportional, it will be a recrystallization solvent and other impurities with which the products were mixed.
Automobile battery acid is 38% H2SO4 and has a destiny of 1.29g/ml. Calculate the molality and the molarity of this solution.
Answer:
[tex]M=5.0M\\\\m=6.2m[/tex]
Explanation:
Hello,
In this case, 38 % is commonly a by mass concentration, meaning that we have 38 grams of solute (sulfuric acid) per 100 grams of solution (water+sulfuric acid):
[tex]38\%=\frac{m_{H_2SO_4}}{m_{H_2SO_4}+m_{H_2O}}[/tex]
Hence, we compute the moles of sulfuric acid in 38 grams by using its molar mass (98 g/mol):
[tex]n_{H_2SO_4}=38g*\frac{1mol}{98g}=0.39mol H_2SO_4[/tex]
Next, the volume of the solution in litres by using the density of the solution:
[tex]V_{solution}=100g*\frac{1mL}{1.29g}*\frac{1L}{1000mL} =0.0775L[/tex]
This is done since the molarity is defined as the ratio of the moles of the solute to the volume of the solution in litres, thus we have:
[tex]M=\frac{n}{V}=\frac{0.38mol}{0.0775L}=5.0M[/tex]
On the other hand, the molality is defined as the ratio of the moles of the solute to the mass of the solvent in kilograms, thus, we compute the mass of water (solvent) as shown below:
[tex]m_{H_2O}=100g-38g=62g*\frac{1kg}{1000g}=0.062kg[/tex]
So compute the molality:
[tex]m=\frac{n_{solute}}{m_{solvent}}=\frac{0.39mol}{0.062kg}=6.2m[/tex]
Regards.
1. The molarity of the solution is 5 M
2. The molality of the solution is 6.26 M
Let the mass of the solution be 100 g.
Therefore, the mass of 38% of H₂SO₄ in the solution is 38 g.
Next, we shall determine the mole of 38 g of H₂SO₄.
Mass of H₂SO₄ = 38 gMolar mass of H₂SO₄ = (2×1) + 32 + (16×4) = 98 g/mol Mole of H₂SO₄ =?Mole = mass / molar mass
Mole of H₂SO₄ = 38 / 98
Mole of H₂SO₄ = 0.388 mole
Next, we shall determine the volume of the solution
Mass of solution = 100 gDensity of solution = 1.29 g/mLVolume of solution =?
Volume = mass / density
Volume of solution = 100 / 1.29
Volume of solution = 77.52 mL
1. Determination of the molarity of the solution
Mole of H₂SO₄ = 0.388 mole Volume of solution = 77.52 mL = 77.52/1000 = 0.07752 LMolarity =?
Molarity = mole / Volume
Molarity = 0.388 / 0.07752
Molarity = 5 M
2. Determination of the molality of the solution
Mole of H₂SO₄ = 0.388 mole Mass of H₂SO₄ = 38 gMass of solution = 100 gMass of water = 100 – 38 = 62 g Mass of water = 62 / 1000 = 0.062 KgMolality =?Molality = mole / mass (Kg) of water
Molality = 0.388 / 0.062
Molality = 6.26 M
Learn more about concentration of solution:
https://brainly.com/question/12154684
Just as carbon dating is used to measure the age of organic material, Argon-40 can be used to measure the age of rocks. A volcanic eruption melts a large chunk of rock, and all gasses are expelled. After cooling, Argon-40 accumulates from the ongoing decay of potassium-40 in the rock (t_1/2 = 1.25E9 years). When a piece of rock is analyzed, it is found to contain 1.38 mmol of potassium-40 and 1.14 mmol of Argon-40. How long did the rock cool?
Answer:
3.77 mg of K-40 decayed into Ar-40.
Data:
1) K-40, Ca-40, Ar-40: all three have the same atomic mass
2) 90% of the potassium-40 will decay into calcium-40
3) 10% of the potassium-40 will decay into argon-40.
4) K-40 inside the rock = 0.81 mg
5) Ar-40 trapped = 0.377 mg
Soltuion:
1) 0.377 mg of Ar-40 is the 10% of the mass of the K-40 that decayed
=> x * 10% = 0.377 mg => x * 0.1 = 0.377mg
=> x = 0.377 mg / 0.1 = 3.77 mg
That means that 3.77 mg of K-40 decayed into Ar-40. And this is the answer to the question.
Additionaly, you can analyze the content of all K-40 and Ca-40, to understand better the case.
2) The mass of the K-40 that decayed into Ca-40 is 9 times (ratio 9:1) the amount that decayed into Ar-40 =>
mass of K-40 that decayed into Ca-40 = 9 * 0.377 = 3.393 mg
3) Total amount of K-40 that decayed = amount that decayed into Ar-40 + amount that decayed into Ca-40 = 0.377mg + 3.393mg = 3.77 mg
4) Original amount of K-40 = amount of K-40 that decayed + amount of K-40 present in the rock = 3.77mg + 0.81 mg = 4.58 mg
5) amount of K-40 that decayed into Ar-40 as percent
% = [3.77 mg / 4.58mg] * 100 = 82.31 %.
