Answer:
NaCl, Na⁺,Cl⁻.
MgCl₂, Mg²⁺, Cl⁻.
CaO, Ca²⁺, O²⁻.
Li₃P, Li⁺, P³⁻.
Al₂S₃, Al³⁺, S²⁻.
Ca₃N₂, Ca²⁺, N³⁻.
FeCl₃, Fe³⁺, Cl⁻.
FeO, Fe²⁺, O²⁻.
Cu₂S, Cu⁺, S²⁻.
Cu₃N₂, Cu²⁺, N³⁻.
ZnO, Zn²⁺, O²⁻.
Ag₂S, Ag⁺, S²⁻.
K₂CO₃, K⁺, CO₃²⁻.
NaNO₃, Na⁺, NO₃⁻.
Ca(HCO₃)₂, Ca²⁺, HCO₃⁻.
Al(OH)₃, Al³⁺,OH⁻.
Li₃PO₄, Li⁺, PO₄³⁻.
K₂SO₄, K⁺, SO₄²⁻.
Explanation:
Sodium chloride. NaCl, formed by the cation Na⁺ and the anion Cl⁻.
Magnesium chloride. MgCl₂, formed by the cation Mg²⁺ and the anion Cl⁻.
Calcium oxide. CaO, formed by the cation Ca²⁺ and the anion O²⁻.
Lithium phosphide. Li₃P, formed by the cation Li⁺ and the anion P³⁻.
Aluminum sulfide. Al₂S₃, formed by the cation Al³⁺ and the anion S²⁻.
Calcium nitride. Ca₃N₂, formed by the cation Ca²⁺ and the anion N³⁻.
Iron(III)chloride. FeCl₃, formed by the cation Fe³⁺ and the anion Cl⁻.
Iron(II)oxide. FeO, formed by the cation Fe²⁺ and the anion O²⁻.
Copper(I)sulfide. Cu₂S, formed by the cation Cu⁺ and the anion S²⁻.
Copper(II)nitride. Cu₃N₂, formed by the cation Cu²⁺ and the anion N³⁻.
Zinc oxide. ZnO, formed by the cation Zn²⁺ and the anion O²⁻.
Silver sulfide. Ag₂S, formed by the cation Ag⁺ and the anion S²⁻.
Potassium carbonate. K₂CO₃, formed by the cation K⁺ and the anion CO₃²⁻.
Sodium nitrate. NaNO₃, formed by the cation Na⁺ and the anion NO₃⁻.
Calcium bicarbonate. Ca(HCO₃)₂, formed by the cation Ca²⁺ and the anion HCO₃⁻.
Aluminum hydroxide. Al(OH)₃, formed by the cation Al³⁺ and the anion OH⁻.
Lithium phosphate. Li₃PO₄, formed by the cation Li⁺ and the anion PO₄³⁻.
Potassium sulfate. K₂SO₄, formed by the cation K⁺ and the anion SO₄²⁻.
An important difference between fuel cells and batteries is that batteries _______. Select the correct answer below: do not require a continuous source of fuel require a continuous source of fuel are able to expel products are constantly resupplied with reactants
Answer: the correct option is that batteries (do NOT require a continuous source of fuel).
Explanation:
A battery can be classified as an electrochemical cell that has the ability to produce electric current. They do NOT require a continuous supply of fuel because it contains all the reactants needed to produce electricity. Below are some examples of batteries that are commonly used:
--> Primary battery: This is a single use battery because it can't be recharged. A typical example is the dry cell.
--> Secondary battery: This type of battery can be recharged. They are used as a power source for smartphones, electronic tablets, and automobiles.
A FUEL CELL is known as a device that converts chemical energy into electrical energy. Fuel cells are similar to batteries but require a continuous source of fuel, often hydrogen. They will continue to produce electricity as long as they are constantly resupplied with reactants. Hydrogen fuel cells have been used to supply power for satellites, space capsules, automobiles, boats, and submarines.
Answer:
accumulate reaction byproducts
Explanation:
Batteries accumulate reaction byproducts. Fuel cells are similar to batteries but require a continuous source of fuel, often hydrogen. They will continue to produce electricity as long as fuel is available. Hydrogen fuel cells have been used to supply power for satellites, space capsules, automobiles, boats, and submarines. A battery is an electrochemical cell or series of cells that produces an electric current.
Good evening everyone Write any three difference between symbol and molecular formula answer it ASAP thank u
Answer:
A molecular formula consists of the chemical symbols for the constituent elements followed by numeric subscripts describing the number of atoms of each element present in the molecule.
For each of the sites specified in the molecules, select whether the site is nucleophilic, electrophilic, or neither. Compound A: The indicated site is a carbon in cyclohexane which is bonded to a bromine and a hydrogen. The indicated carbon in compound A is nucleophilic. neither electrophilic nor nucleophilic. electrophilic. Compound B: The indicated site is the double bond in cyclohexene, a 6 carbon ring with an internal alkene. The indicated bond in compound B is nucleophilic. electrophilic. neither electrophilic nor nucleophilic. Compound C: The indicated site is a carbon double bonded to oxygen, and bonded to O C H 3 and ethyl. The indicated carbon in compound C is neither electrophilic nor nucleophilic. nucleophilic. electrophilic. Compound D: THe indicated site is a carbon bonded to a methyl, two hydrogens and a carbon. There is a nitrogen atom two bonds away. The indicated carbon in compound D is neither electrophilic nor nucleophilic. electrophilic. nucleophilic. Compound E: The indicated site is an oxygen bonded to a carbon and a hydrogen. The indicated oxygen in compound E is neither electrophilic nor nucleophilic. electrophilic. nucleophilic.
The nature of attack on sites in a molecule depends on the nature of such sites. The following are the nature of the sites mentioned in the question:
1) The indicated carbon in compound A is electrophilic.
2) The indicated bond in compound B is nucleophilic.
3) The indicated carbon in compound C is electrophilic.
4) The indicated carbon in compound D is neither electrophilic nor nucleophilic.
5) The indicated oxygen in compound E is nucleophilic.
The terms "electrophilic" and "nucleophilic" are very common in chemistry.
An electrophilic center is usually positively charged, has a positive dipole or is electron deficient hence it attacks negative centers. The term itself means "electron loving". That actually means that it has an affinity for negative charges.
The -I inductive effect of the bromine atom in the carbon in compound A makes that carbon atom to be electrophilic. Also, the carbonyl bond and the O C H 3 attached to the carbon in compound C also makes it electrophilic.
The term "nucleophilic" literately means "nucleus loving". That means a specie that has a high affinity for positive charges. This specie must be electron rich.
The carbon atom in compound B has a double bond which is electron rich and can attack any positive center hence it is nucleophilic. Also, the oxygen atom in E bears two lone pairs of electrons which can attack any positive center in a molecule hence the oxygen atom is also nucleophilic.
In compound D, the carbon atom is bonded to a methyl, two hydrogens and a carbon. There is a nitrogen atom two bonds away. There is no +I or -I inductive effect on this carbon atom because the nitrogen atom is far away. Therefore, the indicated carbon in compound D is neither electrophilic nor nucleophilic.
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Draw the structure of the organic product(s) of the Grignard reaction between methyl benzoate and excess phenylmagnesium bromide, followed by aqueous workup. You do not have to consider stereochemistry. If a compound is formed more than once, add another sketcher and draw it again. Alternatively, you may use the square brackets tool to add stoichiometries greater than one. Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. Separate multiple products using the sign from the drop-down menu.
Answer:
See explanation and image attached
Explanation:
The product of the Grignard reaction between methyl benzoate and excess phenylmagnesium bromide is triphenyl methanol.
The reaction proceeds by nucleophillic reaction as the carbonyl moiety is attacked. A tetrahedral intermediate is formed. Loss of the -OMe group is accompanied by the attack of the first molecule of PhMgBr.
Attack by a second PhMgBr molecule yields trimethyl phenoxide. Protonation of this specie yields the final product which is obtained by aqueous workup.
(d) 40g of sulphur
Calculate the number of moles of 40g of sulphur
Answer:
It is 1.25 moles
Explanation:
Molar mass of sulphur = 32 g
[tex]{ \bf{moles = \frac{given \: mass}{molar \: mass} }} [/tex]
Substitute:
[tex]{ \sf{moles = \frac{40}{32} }} \\ { \sf{ = 1.25 \: moles}}[/tex]
The number of organic compounds that have been reported in 2019 is more than 10 million. This is due to carbon's ability to bond to other carbon molecules, called
electron affinity.
ionic bonding.
catenation.
electronegativity.
Answer:
catenation
Explanation:
Carbon atoms have four electrons to share in bonding environments to get to the ideal octet. To do this, it bonds with other carbon molecules, called catenation. Catenation is the ability of an atom to bond and share electrons with other atoms of its kind.
The specific heat capacity of lead is 0.13 J/g-K. How much heat (in J) is required to raise the temperature of 15 g of lead from 22 °C to 37 °C? a. 5.8 × 10-4 J b. 0.13 J c. 29 J d. 2.0 J e. -0.13 J
Answer:
c. 29 J
Explanation:
Step 1: Given data
Specific heat capacity of Pb (c): 0.13 J/g.K (= 0.13 J/g.°C)Mass of Pb (m): 15 gInitial temperature: 22 °CFinal temperature: 37 °CStep 2: Calculate the temperature change
ΔT = 37 °C - 22 °C = 15 °C
Step 3: Calculate the heat (Q) required to raise the temperature of the lead piece
We will use the following expression.
Q = c × m × ΔT
Q = 0.13 J/g.°C × 15 g × 15 °C = 29 J
14. What is the oxidation number of oxygen in HSO4 -
Answer:
2
Explanation:
i did this
150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reaction is completed; calculate the concentration of nitrate ions in solution. Assume that the total volume of the solution is 3.0 x 10^2 mL
Answer:
[tex]0.175\; \rm mol \cdot L^{-1}[/tex].
Explanation:
Magnesium chloride and silver nitrate reacts at a [tex]2:1[/tex] ratio:
[tex]\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)[/tex].
In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:
[tex]\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}[/tex].
The precipitate silver chloride [tex]\rm AgCl[/tex] is insoluble in water and barely ionizes. Hence, [tex]\rm AgCl\![/tex] isn't rewritten as ions.
Net ionic equation:
[tex]\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}[/tex].
Calculate the initial quantity of nitrate ions in the mixture.
[tex]\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}[/tex].
Since nitrate ions [tex]\rm {NO_3}^{-}[/tex] do not take part in any reaction in this mixture, the quantity of this ion would stay the same.
[tex]n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol[/tex].
However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be [tex](1/2)[/tex] of the concentration in the original solution.
[tex]\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
During the course of your experiment you have obtained the following data: mass of the hydrate: 1.973 g mass of the anhydrate: 1.196 g The formula of the anhydrous salt: CaCl2 Calculate the following: (round to correct the number of significant figures and include units as required). mass of water loss in . number of moles of anhydrous salt after heating, in moles number of moles of water lost, in moles • number of moles of water per mole of hydrate, in moles (round to the whole number) provide the formula of a hydrate Note: you will not be able to add the bscript and leave one space between ionic compound and water.
Answer:
See explanation
Explanation:
Mass of water lost = mass of hydrated salt - mass of anhydrous salt
Mass of water lost = 1.973 g - 1.196 g = 0.777g
Number of moles of water lost = 0.777g/18g/mol = 0.043 moles
Number of moles of anhydrous salt = 1.196 g /111g/mol = 0.011 moles
To obtain the number of moles of water of crystalization per hydrate molecule;
Number of moles of anhydrous salt = number of moles of hydrated salt
0.011 = 1.973 /111 + 18x
0.011(111 + 18x) = 1.973
1.221 + 0.198x = 1.973
0.198x = 1.973 - 1.221
x= 4
Hence, there are 4 moles of water per hydrate molecule. The formula of the hydrate is CaCl2.4H2O
Let's assume you were given 2.0 g benzil, 2.2 g dibenzyl ketone, 50 mL 95% ethanol and 0.3 g potassium hydroxide to synthesize tetraphenylcyclopentadienone. You isolated 3.0 g of tetraphenylcyclopentadienone. What is the % yield
Answer:
the % yield is 82%
Explanation:
Given the data in the question,
we know that;
Molar mass of benzil is 210.23 g·mol−1
Molar mass of dibenzyl ketone is 210.27 g·mol−1
Molar mass of tetraphenylcyclopentadienone is 384.5 g·mol−1
Now,
2.0 g benzil = 2 g / 210.23 g·mol−1 = 0.0095 mole
2.2 g dibenzyl ketone = 2.2 / 210.27 = 0.0105 mole
3.0 g of tetraphenylcyclopentadienone = 3 / 384.5 = 0.0078 mole
Now, the limiting reagent is benzil. 0.0095 mole can reacts wiyh 0.0095 mole of dibenzyl ketone
percentage yield = ( 0.0078 mole / 0.0095 mole ) × 100%
= 0.82 × 100%
= 82%
Therefore, the % yield is 82%
Forcus on the yellow highlighted texts, your help is appreciated.
[tex]{ \sf{ \red{no \: pranks}}}[/tex]
Answer:
Transition temperature is the temperature at which a substance changes from one state to another.
Allotropy is the existence of an element in many forms.
2. Calculate the wavelength of the emitted photon from hydrogen for the transition from ni = 3 to nf = 2. What part of the visible spectrum is this wavelength? Visible wavelengths are: Red 700 - 620 nm, Yellow 620 - 560 nm, Green 560 - 500 nm, Blue 500 - 440 nm, and Violet 440 - 400 nm.
Answer:
The correct answer is "654.54 nm".
Explanation:
According to the question,
⇒ [tex]\frac{1}{\lambda}= Rh(\frac{1}{n1^2} -\frac{1}{n2^2} )[/tex]
By substituting the values, we get
[tex]=1.1\times 10^7(\frac{1}{4} -\frac{1}{9} )[/tex]
[tex]=1.1\times 10^7(\frac{9-4}{36} )[/tex]
[tex]=1.1\times 10^7(\frac{5}{36} )[/tex]
[tex]=654.54\ nm[/tex]
Thus the above is the right solution.
Calculate the mass of isoborneol in 2.5 mmol of isoborneol and the theoretical yield (in grams) of camphor from that amount of isoborneol
isoborneol = 154.25 g mol?1
Camphor, Molar mass = 152.23 g/mol
Answer:
[tex]m_{isoborneol }=0.39g\\\\m_{Camphor}=0.38g\\[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to infer that the reaction whereby isoborneol goes to camphor occurs in a 1:1 mole ratio, that is why the theoretical yield of the latter is also 2.5 mmol (0.0025 mol) but the masses can be calculated as follows:
[tex]m_{isoborneol }=0.0025mol*\frac{154.25g}{1mol} =0.39g\\\\m_{Camphor}=0.0025mol*\frac{152.23 g}{1mol} =0.38g\\[/tex]
Because of the fact this is a rearrangement reaction whereas the number of atoms is not significantly modified.
Regards!
The enthalpy of formation for CO2 (s) and CO2 (g) is: -427.4 KJ/mole and -393.5 KJ/mole, respectively. The sublimation of dry ice is described by CO2 (s) → CO2 (g).
The enthalpy needed to sublime 986 grams of CO2 is:
(a) 181.5 Kcal
(b) 611.7 Kcal
(c) 248.3 Kcal
(d) 146.2 Kcal
Answer:
a. 181.5 kcal
Explanation:
Step 1: Calculate the enthalpy of the process (ΔH°).
Let's consider the following process.
CO₂(s) → CO₂(g)
We can calculate the enthalpy of the process using the following expression.
ΔH° = ∑ np × ΔH°f(p) - ∑ nr × ΔH°f(r)
ΔH° = 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CO₂(s))
ΔH° = 1 mol × (-393.5 kJ/mol) - 1 mol × (-427.4 kJ/mol) = 33.9 kJ
According to the balanced equation, 33.9 kJ are required to sublime 1 mole of CO₂.
Step 2: Convert 986 g of CO₂ to moles
The molar mass of CO₂ is 44.01 g/mol.
986 g × 1 mol/44.01 g = 22.4 mol
Step 3: Calculate the enthalpy needed to sublime 22.4 moles of CO₂
22.4 mol × 33.9 kJ/1 mol = 759 kJ
We can convert it to Kcal using the conversion factor 1 kcal = 4.184 kJ.
759 kJ × 1 kcal/4.184 kJ = 181.5 kcal
What is the energy of a photon emitted with a wavelength of 654 nm?
O A. 3.04 x 10^-19 J
O B. 1.01 * 10^-27 J
O C. 1.30 x 10^-22 J
O D. 4.33 * 10^-22 J
Answer:
A. 3.04×10^-19J
Explanation:
Hope this will help you.
what gasous product would you expect when water is drop over calcium carbide
Answer:
Ethyn gas ( acetylene gas )
Explanation:
All group II carbides react with water to form ethyn gas apart from beryllium which produces methane gas.
[tex]{ \sf{CaC_{2(s)} + 2H _{2} O_{(l)} → Ca(OH) _{2(s)} + C _{2} H _{2(g)} }}[/tex]
Too many objects inside a laboratory fume hood can disrupt the airflow and possibly compromise you safety. Which of the following are considered best practices in the use of a laboratory fume hood?
a. Open the sash as much as possible
b. Work at least 25 cm inside the hood
c. Use fast, quick movements to limit your exposure
d. Place objects to one side—work on other side
e. Use a raised along the back of the hood
Best practices for fume hoods: work 25 cm inside, organize items to one side, use raised ledge; avoid open sash and quick movements.
Laboratory fume hoods must be used safely. Workers should operate at least 25 cm within the hood to preserve ventilation and avoid dangerous chemicals. Place things on one side of the hood to preserve ventilation and prevent clogging.
A raised ledge on the rear of the hood prevents things from falling in and impeding airflow. Avoid fully opening the sash to maintain ventilation and containment. Fast, rapid motions can interrupt airflow, so prevent them. These practises guarantee the fume hood contains harmful compounds, making the lab safer. Therefore, option (B), (D) and (E) are correct.
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Calculate the amount of water (in grams) that must be added to (a) 6.80 g of urea [(NH2)2CO] in the preparation of a 9.95 percent by mass solution: g (b) 29.3 g of MgBr2 in the preparation of a 1.70 percent mass solution: g
Explanation:
Amount of water required in each case:
(a)The mass% of the solution is:9.95
Mass of solute that is urea is 6.80 g
To determine the mass of solvent water use the formula:
[tex]mass percent=\frac{mass of solute}{mass of solution} x 100\\\\9.95=(6.80g/mass of solution )x100\\mass of solution =(6.80 /9.95)x100\\=68.3 g[/tex]
Hence the mass of solvent = mass of solution - the mass of solute
=68.3 g - 6.80g
=61.5 g
Hence, the answer is mass of solvent water required is 61.5 g.
(b) Given mass%=1.70
mass of solute MgBr2 = 29.3 g
The mass of solvent water required can be calculated as shown below:
[tex]mass percent=\frac{mass of solute}{mass of solution} x 100\\\\1.70=(29.3g/mass of solution )x100\\mass of solution =(29.3 g /1.70)x100\\=1720 g[/tex]
The mass of the solution is 1720 g.
Mass of solvent water = mass of solution - mass of solute
=1720 g - 29.3 g
=1690.7 g
Answer: The mass of water required is 1690.7 g.
Suppose that you add 24.3 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 3.14 oC compared to pure benzene. What is the molar mass (in g/mol) of the unknown compound
Solution :
We know that :
[tex]$\Delta T_f = k_f.m$[/tex] and [tex]$m=\frac{w_2}{m_2 \times w_1}$[/tex]
Then, [tex]$\Delta T_f = k_f.\frac{w_2}{m_2.w_1}$[/tex] ..................(1)
Where,
[tex]w_1[/tex] = amount of solvent (in kg)
[tex]w_2[/tex] = amount of solute (in kg)
[tex]m_2[/tex] = molar mass of solute (g/mole)
[tex]m[/tex] = molality of solution (mole/kg)
Given :
[tex]\Delta T_f[/tex] = [tex]3.14\ ^\circ C[/tex], [tex]k_f= 5.12\ ^\circ C/m[/tex]
[tex]=5.12 \ ^\circ C/mole/kg[/tex]
[tex]=5.12 \ ^\circ C \ kg/mole[/tex]
[tex]w_1[/tex] = 0.250 kg, [tex]w_2[/tex] = 24.3 g
Then putting this values in the equation is (1),
[tex]$3.14 = \frac{5.12 \times 24.3}{m_2 \times 0.250}$[/tex]
[tex]$m_2 = \frac{5.12 \times 24.3}{3.14 \times 0.250}$[/tex]
[tex]m_2= 158.49[/tex] g/mole
So, the molar mass of the unknown compound is 158.49 g/mole.
What is the maximum mass of PH3 that can be formed when 62.0g of phosphorus reacts with
4.00g of hydrogen?
P4(g)+ 6H2(g) → 4PH3(g)
Answer: The mass of [tex]PH_3[/tex] produced is 45.22 g
Explanation:
Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.
Excess reagent is defined as the reagent which is left behind after the completion of the reaction.
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
For [tex]P_4[/tex]:Given mass of [tex]P_4[/tex] = 62.0 g
Molar mass of [tex]P_4[/tex] = 124 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }P_4=\frac{62.0g}{124g/mol}=0.516mol[/tex]
For [tex]H_2[/tex]:Given mass of [tex]H_2[/tex] = 4.00 g
Molar mass of [tex]H_2[/tex] = 2 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }H_2=\frac{4.0g}{2g/mol}=2mol[/tex]
The chemical equation follows:
[tex]P_4(g)+6H_2(g)\rightarrow 4PH_3(g)[/tex]
By stoichiometry of the reaction:
If 6 moles of hydrogen gas reacts with 1 mole of [tex]P_4[/tex]
So, 2 moles of hydrogen gas will react with = [tex]\frac{1}{6}\times 2=0.333mol[/tex] of [tex]P_4[/tex]
As the given amount of [tex]P_4[/tex] is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Thus, hydrogen gas is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 6 moles of [tex]H_2[/tex] produces 4 mole of [tex]PH_3[/tex]
So, 2 moles of [tex]H_2[/tex] will produce = [tex]\frac{4}{6}\times 2=1.33mol[/tex] of [tex]PH_3[/tex]
We know, molar mass of [tex]PH_3[/tex] = 34 g/mol
Putting values in equation 1, we get:
[tex]\text{Mass of }PH_3=(1.33mol\times 34g/mol)=45.22g[/tex]
Hence, the mass of [tex]PH_3[/tex] produced is 45.22 g
The reaction responsible for producing the heat that maintains the temperature of your body is
Answers
A.
metabolism.
B.
catabolism.
C.
anabolism.
D.
photosynthesis.
Question
Answer:
A
Explanation:
it increase the rate of reaction when necessary
The chemical change that is responsible for producing the heat that maintains the temperature of your body is metabolism.
What is chemical change?
Chemical changes are defined as changes which occur when a substance combines with another substance to form a new substance.Alternatively, when a substance breaks down or decomposes to give new substances it is also considered to be a chemical change.
There are several characteristics of chemical changes like change in color, change in state , change in odor and change in composition . During chemical change there is also formation of precipitate an insoluble mass of substance or even evolution of gases.
There are three types of chemical changes:
1) inorganic changes
2)organic changes
3) biochemical changes
During chemical changes atoms are rearranged and changes are accompanied by an energy change as new substances are formed.
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Nitric acid and nitrogen monoxide react to form nitrogen dioxide and water, like this: At a certain temperature, a chemist finds that a 7.7 L reaction vessel containing a mixture of nitric acid, nitrogen monoxide, nitrogen dioxide, and water at equilibrium has the following composition: compound amount
HNO 16.2 g 11.0 g 18.6 g H20 236.7 g 3 NO NO
Calculate the value of the equilibrium constant K for this reaction. Round your answer to 2 significant digits.
Answer:
K = 3.3
Explanation:
Nitric acid, HNO3, reacts with nitrogen monoxide, NO, to produce nitrogen dioxide, NO2 and water H2O as follows:
2HNO3(g) + NO(g) → 3NO2(g) + H2O(g)
Where equilibrium constant, K, is:
K = [NO2]³[H2O] / [HNO3]²[NO]
[] is the molar concentration of each species at equilibrium.
To solve this question we need to find molarity of each gas and replace these in the equation as follows:
[NO2] -Molar mass NO2-46.0g/mol-
18.6g * (1mol/46.0g) = 0.404mol / 7.7L = 0.0525M
[H2O] -Molar mass:18.01g/mol-
236.7g * (1mol/18.01g) = 13.14 moles / 7.7L = 1.707M
[HNO3] -Molar mass:53.01g/mol-
16.2g * (1mol/53.01g) = 0.3056 moles / 7.7L = 0.0397M
[NO] -Molar mass: 30.0g/mol-
11.0g * (1mol/30.0g) = 0.367 moles / 7.7L = 0.0476M
Replacing:
K = [NO2]³[H2O] / [HNO3]²[NO]
K = [0.0525M]³[1.707M] / [0.0397M]²[0.0476M]
K = 3.3
A substance is tested and has a pH of 7.0. How would you classify it?
Oxygen is composed of three isotopes: oxygen-16, oxygen-17 and oxygen-18 and has an average atomic mass of 15.9982 amu. Oxygen-17 has a mass of 16.988 amu and makes up 0.032% of oxygen. Oxygen-16 has a mass of 15.972 amu and oxygen-18 has a mass of 17.970 amu. What is the percent abundance of oxygen-18?
Answer:
The percent abundance of oxygen-18 is 1.9066%.
Explanation:
The average atomic mass of oxygen is given by:
[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]
Where:
m: is the atomic mass
%: is the percent abundance
Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:
[tex] 1 = \%_{16} + \%_{17} + \%_{18} [/tex]
[tex] 1 = x + 3.2 \cdot 10^{-4} + \%_{18} [/tex]
[tex] \%_{18} = 1 - x - 3.2 \cdot 10^{-4} [/tex]
Hence, the percent abundance of O-18 is:
[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]
[tex]15.9982 = 15.972*x + 16.988*3.2 \cdot 10^{-4} + 17.970*(1 - 3.2 \cdot 10^{-4} - x)[/tex]
[tex] x = 0.980614 \times 100 = 98.0614 \% [/tex]
Hence, the percent abundance of oxygen-18 is:
[tex]\%_{18} = (1 - 3.2 \cdot 10^{-4} - 0.980614) \times 100 = 1.9066 \%[/tex]
Therefore, the percent abundance of oxygen-18 is 1.9066%.
I hope it helps you!
Suppose a 250.mL flask is filled with 1.7mol of H2 and 0.90mol of I2. The following reaction becomes possible:
+H2gI2g 2HIg
The equilibrium constant K for this reaction is 5.51 at the temperature of the flask.
Calculate the equilibrium molarity of I2. Round your answer to two decimal places.
Explanation:
here's the answer. I just plug the expression into my calculator and find the intercept to avoid the quadratic formula
Why is bromine more electronegative than iodine?
Answer
Accordingly the order of electronegativity of the given elements would be: Fluorine > Chlorine > Bromine > Iodine. ( Fluorine has the highest electronegativity.)
A mixture of methane and carbon dioxide gases contains methane at a partial pressure of 431 mm Hg and carbon dioxide at a
partial pressure of 504 mm Hg. What is the mole fraction of each gas in the mixture?
XCHA
Xc02
Answer:
XCH₄ = 0.461
XCO₂ = 0.539
Explanation:
Step 1: Given data
Partial pressure of methane (pCH₄): 431 mmHgPartial pressure of carbon dioxide (pCO₂): 504 mmHgStep 2: Calculate the total pressure in the container
We will sum both partial pressures.
P = pCH₄ + pCO₂
P = 431 mmHg + 504 mmHg = 935 mmHg
Step 3: Calculate the mole fraction of each gas
We will use the following expression.
Xi = pi / P
XCH₄ = pCH₄/P = 431 mmHg/935 mmHg = 0.461
XCO₂ = pCO₂/P = 504 mmHg/935 mmHg = 0.539
A reaction vessel for synthesizing ammonia by reacting nitrogen and hydrogen is charged with 6.54 kg of H2 and excess N2. A total of 30.4 kg of NH3 are produced. What is the percent yield of the reaction
Explanation:
The given data is:
The mass of hydrogen is 6.54 kg.
The actual yield is 30.4 kg.
The balanced chemical equation of the reaction is:
[tex]N_2(g)+3H_2(g)<=>2NH_3(g)[/tex]
At first the theoretical yield should be calculated by using the balanced chemical equation:
3 mol. of hydrogen forms ---- 2 mol. of ammonia.
The molar mass of hydrogen is 2.0 g/mol.
The molar mass of ammonia is 17.0 g/mol.
Hence, the above statement can be rewritten as:
6g of hydrogen forms --- 34g of ammonia.
Then,
6.54g of hydrogen forms :
[tex]6.54 kg x 34 g / 6 g\\=37.1 kg[/tex]
% yield = (actual yield /theoretical yield )x 100
=(30.4 kg /37.1 kg )x100
=81.9
Hence, % yield is 81.9.
If a hydrogen of an alkane is replaced by NH, the compound becomes_________
a. alcohol
b. carboxylic acid
c. phenol
d. amine
Answer:
d. amine.
It becomes an amine.
Explanation:
With general formular
[tex]{ \bf{primary \: amine :R - NH _{2}}} \\ { \bf{secondary \: amine : R {}^{i} - NH - R}} \\ { \bf{tertiary \: amine :R {}^{ii} - N(R {}^{i} ) - R }}[/tex]
R is the aryl group such as alkane
