Al2O3 would be typed in as Al2O3 Al2(SO4)3 would be typed in as Al2(SO4)3.

Required:
a. Write the formula for calciumnitride.
b. Write the formula for magnesium phosphide
c. Write the formula for rubidium chromate
d. Write the formula for aluminum nitrate
e. Write the formula for ammonium arsenide
f. Write the formula for nickel(II) nitrite
g. Write the formula for copper(I) sulfate
h. Write the formula for iron(III) nitrate
i. Write the formula for manganese(II) nitride

Answers

Answer 1

Answer:

Explanation:

Hello,

We are required to write the chemical formula of the following compounds

1. Calcium nitride = Ca₃N₂

2. Magnesium Phosphide = Mg₃P₂

3. Rubidium Chromate = Rb₂CrO₄

4. Aluminium nitrate = Al(NO₃)₃

5. Ammonium Arsenide = (NH₄)₃As

6. Nickel(ii) nitrite = Ni(NO₂)₂

7. Copper(i)sulfate = Cu₂SO₄

8. Iron(iii)nitrate = Fe(NO₃)₃

9. Manganese(ii)nitrate = Mn(NO₃)₂

The left hand side are the common names when the right hand side are the chemical formula.


Related Questions

Which phrase best describes heat?
A. the energy that an object has as a result of its temperature
B
the average translational kinetic energy of the particles in an object
C
the energy transferred between objects at different temperatures
D
the total amount of energy possessed by the particles in an object

Answers

Answer: A

Explanation: Heat is the temperature of an object.

Answer:

Option C

the energy transferred between objects at different temperature.

Suppose of potassium iodide is dissolved in of a aqueous solution of silver nitrate. Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the potassium iodide is dissolved in it.

Answers

The given question is incomplete, the complete question is:

Suppose 1.27 g of potassium iodide is dissolved in 100. mL of a 44.0 m M aqueous solution of silver nitrate. Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the potassium iodide is dissolved in it. Round your answer to 3 significant digits.

Answer:

The correct answer is 0.0325 M.

Explanation:

The mass of potassium iodide or KI mentioned in the question is 1.27 grams, the molar mass of KI is 166 g/mol. The formula for determining the no of moles of the substance is mass/molar mass. Thus, the moles of KI in 1.27 grams will be,  

= 1.27g / 166 g/mol = 0.00765 moles.  

KI = K⁺ + I⁻

Therefore, the moles of KI will be equivalent to moles of iodide anion, that is, 0.00765 moles.  

The moles of silver nitrate or AgNo3 in the solution can be determined by using the formula, molarity (M) * volume in liters. The molarity of silver nitrate given in the question is 44 mM and the volume used is 100 ml or 100/1000 L. Now putting the values we get,  

= (44 M/1000) * (100 L/1000) = 0.0044 moles

The moles of silver nitrate is equivalent to moles of silver ion, which is further equivalent to the moles of iodide ion that has taken part in precipitation = 0.0044 moles.  

The moles left of I⁻ in the solution will be,  

0.00765 - 0.0044 = 0.00325

Now, the final molarity of iodide ion in the solution will be,  

= moles/volume in liters

= 0.00325 moles / 0.100 L = 0.0325 M

The functional groups in an organic compound can frequently be deduced from its infrare d absorption spectrum. A compound, C3H6O2, exhibits intense absorption at 1740 cm-1, accompanied by a band at 1050 cm-1. No absorption above 3000 cm-1 is present .
What functional class(es) does the compound belong to?
List only classes for which evidence is given here. Attach no significance to evidence not cited explicitly.
Do not over-interpret exact absorption band positions. None of your inferences should depend on small differences like 10 to 20 cm-1.
The functional class(es) of this compound is(are) .(Enter letters from the table below, in any order, with no spaces or commas.)
a. alkane (List only if no other functional class applies.)
b. alkene
c. terminal alkyne
d. internal alkyne
e. arene
f. alcohol
g. ether
h. amine
i. aldehyde or ketone
j. carboxylic acid
k. ester
l. nitrile

Answers

Answer:

(b.) alkene

(i) aldehyde/ketone

(k.) ester

Explanation:

Peaks observed at:

1740 cm-1 indicates the presence of a carbonyl group: C=O aldehyde, ketone, esters.1050 cm-1 shows presence of carbo bonded to electronegative elements e.g. C-N or C-O3000cm-1 is usually indicative of alcohols or carboxylic acid -OH group, which rules out those classes.

An increase in the temperature of reactant causes an increase in the rate of reaction.
Which of the following is the best explanation?

a)the concentration of reactant increases
b) the activation energy decreases
c) the collision frequency increases
d) the fraction of collision with total kinetic energy larger than activation energy
increases

Answers

Answer:

d) The fraction of collision with total kinetic energy larger than activation energy  increases.

Explanation:

Hello,

In this case, kinetic models explain how the rate of a chemical reaction is affected by several factors. In such a way, specifically for temperature, when it increases, the average velocity of the particles is also increased, for that reason, the collision frequency increases since the molecules are more likely to collide as they move faster and encounter to each other.

Nonetheless, it is the minor reason because the main reason is that the effective collisions increase when the temperature is increased, and they are related with the fraction of collision with total kinetic energy that turns out larger than the activation energy, therefore, answer is d).

Best regards.

A solution is made by mixing 10.2 grams of CaCl2 in 250 grams of water. What is the molality of the

solution?

Answers

Answer:

The molality of this solution is 0.368 molal

Explanation:

Step 1: Data given

Mass of CaCl2 = 10.2 grams

Molar mass of CaCl2 = 110.98 g/mol

Mass of water = 250 grams = 0.250 kg

Step 2: Calculate the number of moles CaCl2

Moles CaCl2 = mass CaCl2 / molar mass CaCl2

Moles CaCl2 = 10.2 grams / 110.98 g/mol

Moles CaCl2 = 0.0919 moles

Step 3: Calculate the molality of the solution

Molality solution = moles CaCl2 / mass water

Molality solution = 0.0919 moles / 0.250 kg

Molality solution = 0.368 mol / kg = 0.368 molal

The molality of this solution is 0.368 molal

Answer:

0.367 m

Explanation:

We have to start with the molality equation:

[tex]m=\frac{mol~of~solute}{Kg~of~solvent}[/tex]

If we want to calculate the molality we need the moles of the solute and the Kg of the solvent. In this case the solute is [tex]CaCl_2[/tex] and the solvent is [tex]H_2O[/tex].

Lets start with the the moles of solute. For this calculation we need the molar mass of [tex]CaCl_2[/tex] ([tex]111~g~CaCl_2[/tex]) and the mass, so:

[tex]10.2~g~CaCl_2\frac{1~mol~CaCl_2}{111~g~CaCl_2}[/tex]

[tex]0.0918~mol~CaCl_2[/tex]

Now, we can calculate the Kg of solvent, if we do the conversion from grams to Kg:

[tex]250~g~CaCl_2\frac{1~Kg}{1000~g}[/tex]

[tex]0.25~Kg[/tex]

With these values we can calculate the "molality":

[tex]m=\frac{0.0918~mol~CaCl_2}{0.25~Kg}[/tex]

[tex]m=0.3672[/tex]

I hope it helps!

Interpret and describe an example of a chemical formula. Summarize the two parts a chemical formula must have.

Answers

Answer and Explanation:

A chemical formula is composed by chemical symbols (in letters, which indicate the chemical elements) and subscripts (numbers). For example, let see the chemical formula for carbon dioxide:

CO₂

The letters C and O are the chemical symbols for the elements carbon (C) and oxygen (O). The number 2 in subscript indicates that there are 2 atoms of O per molecule. The subscript 1 is generally not indicated in chemical formulae, so it is assumed that the number of atoms of C is 1.

Summarizing, the chemical formula CO₂ indicates us that the molecule if formed by 1 atom of the element carbon (C) and 2 atoms of the element oxygen (O).

There is a pattern to the phases of the moon. The time between one full moon to another is about a;new b;waxing crescent c;1st quarter d;full

Answers

Answer: a. new

Explanation:

The time between one full moon to another is new moon. The lunation is the average time taken for the formation of new moon to the next. The new moon is the initial visible crescent of the Moon after being conjugated with the Sun. Per lunar cycle can be assigned with a lunation number for the purpose of identification.

How many grams of mercury would be contained in 15 compact fluorescent light bulbs?

Answers

Answer:

Grams of mercury= 0.06 g of Hg

Note: The question is incomplete. The complete question is as follows:

A compact fluorescent light bulb contains 4 mg of mercury. How many grams of mercury would be contained in 15 compact fluorescent light bulbs?

Explanation:

Since one fluorescent light bulb contains 4 mg of mercury,

15 such bulbs will contain 15 * 4 mg of mercury = 60 mg

1 mg = 0.001 g

Therefore, 60 mg = 0.001 g * 60 = 0.06 g of mercury.

Compact fluorescent lightbulbs (CFLs) are tubes containing mercury and noble gases. When electricity is passed through the bulb, electron-streams flow from a tungsten-coated coil. They collide with mercury atoms, exciting their electrons and creating flashes of ultraviolet light. A phosphor coating on the inside of the tube absorbs this UV light flashes and re-emits it as visible light. The amount of mercury in a fluorescent lamp varies from 3 to 46 mg, depending on lamp size and age.

Total amount of mercury in 15 compact fluorescent light bulbs is 0.06 gram of mercury.

Compact fluorescent light bulbs and mercury:

What information do we have?

Number of compact fluorescent light bulbs = 15 bulb

Amount of mercury in each bulb = 4 mg

Total amount of mercury = Number of compact fluorescent light bulbs × Amount of mercury in each bulb

Total amount of mercury = 15 × 4

Total amount of mercury = 60 mg

Total amount of mercury = 60 / 1000

Total amount of mercury = 0.06 gram of mercury

Find out more information about 'Mercury'.

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43. Calculate the equilibrium constant at the temperature given. (a) O2 (g) + 2F2 (g) ⟶ 2F2 O(g) (T = 100 °C) (b) I2 (s) + Br2 (l) ⟶ 2IBr(g) (T = 0.0 °C) (c) 2LiOH(s) + CO2 (g) ⟶ Li2CO3 (s) + H2 O(g) (T = 575 °C) (d) N2 O3 (g) ⟶ NO(g) + NO2 (g) (T = −10.0 °C) (e) SnCl4 (l) ⟶ SnCl4 (g) (T = 200 °C)

Answers

Answer:

Explanation:

O₂(g) + 2F₂(g) ↔ 2F₂(g)

Stabdard ΔG values are

[tex]\Delta G_f[F_2O]=41.9kJ/mol =41900J/mol[/tex]

[tex]\Delta G_f[O_2]=0\\\\ \Delta G_f[F_2]=0[/tex]

[tex]\Delta G^0=\sum \Delta G^\circ (products)- \sum \Delta G ^\circ (reactants)[/tex]

[tex]\Delta G^\circ = [2 \times 41900]-0\\\\=83800J/mol[/tex]

Now,

[tex]\Delta G^\circ =-RTInK[/tex]

Given T = 100°C

= 100+ 273.15 = 373.15K

R = 8.314J/k / mol

so,

83800 = -8.314 * 373.15 * InK

InK = -27.0116

K = 1.858 * 10⁻¹²

Equilibrium constant =  1.858 * 10⁻¹²

100 mL of a buffer that consists of 0.20 M NH3 and 0.20 M NH4Cl is titrated with 25 mL of 0.20 M HCl. Calculate the pH of the resulting solution given that the Kb for NH3 is 1.8 x 10-5.

Answers

Answer:

pH = 9.03

Explanation:

The equilibrium of the NH₄Cl / NH₃ buffer in water is:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

Initial moles of both NH₃ and NH₄⁺ are:

0.100L ₓ (0.20 mol / L) = 0.0200 moles

The NH₃ reacts with HCl producing NH₄⁺, thus:

NH₃ + HCl → NH₄⁺ + Cl⁻

That means, moles of HCl added to the solution are the same moles are consumed of NH₃ and produced of NH₄⁺

Moles added of HCl were:

0.025L ₓ (0.20mol / L) = 0.0050 moles of HCl. Thus, final moles of NH₃ and NH₄⁺ are:

NH₃: 0.0200 moles - 0.0050 moles = 0.0150 moles

NH₄⁺: 0.0200 moles + 0.0050 moles = 0.0250 moles.

Using H-H equation for bases:

pOH = pKb + log [NH₄⁺] / [NH₃]

Where pKb is -log Kb = 4.745.

Replacing:

pOH = 4.745 + log 0.0250mol / 0.0150mol

pOH = 4.967

As pH = 14- pOH

pH = 9.03

A chemistry student needs of ethanolamine for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of ethanolamine is . Calculate the mass of ethanolamine the student should weigh out. Round your answer to significant digits.

Answers

Answer:

mass of ethanolamine 61.2 g

Explanation:

Data:

V (Volume) = 60 mL

ρ (Density)= 1.02 [tex]\frac{g}{cm^{3} }[/tex] = 1.02 [tex]\frac{g}{mL}[/tex]

With the given data and using the density formula:

ρ = [tex]\frac{mass}{volume}[/tex]

It is possible to calculate the grams of ethanolamine that the student should weigh out. The mass in the formula is cleared and the data is replaced.

mass =  ρ * volume

mass = 1.02 [tex]\frac{g}{mL}[/tex] * 60 mL = 61.2 g of ethanolamine

Given that a 15.00 g milk chocolate bar contains 9.500 g of sugar, calculate the percentage of sugar present
in 15.00 g of milk chocolate bar keeping in mind that the answer should have four significant figures (two
decimal places).

Answers

Answer:

63.33%

Explanation:

To find the percentage of sugar in the milk chocolate just divide 15 by 9.5. This will give you 0.63333 (recurring). To find the percentage, times it by 100, which will give you 63.33%.

Hope this helped :)

The percentage of sugar present in the milk would be  63.33 %.

What are significant figures?

In positional notation, significant figures refer to the digits in a number that is trustworthy and required to denote the amount of something, also known as the significant digits, precision, or resolution.

As given in the problem a 15.00 g milk chocolate bar contains 9.500 grams of sugar, and we have calculated the percentage of sugar present,

The percentage of sugar present in the milk = 9.500 ×100/15

                                                                                = 63.33 %

Thus, the percentage of sugar present in the milk would be 63.33 %.

To learn more about significant figures here, refer to the link;

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516 mL of a 3.82 M sodium sulfate (Na2S04) solution is diluted with 0.875 L of water. What is the new concentration in molarity?

Answers

Answer

Molarity = [tex]1.4mol/L[/tex]

Explanation:

Molarity provides the number of moles of solute per liter of solution (moles/Liter). It is a means by which concentration of solution is measured.

SEE THE ATTACHMENT BELOW FOR STEP BY STEP SOLUTION.

Answer:

1.42 M

Explanation:

In this case have a dilution problem, therefore we need to use the dilution equation:

[tex]C_1*V_1=C_2*V_2[/tex]

What values we have?

[tex]C_1=3.82M[/tex]

[tex]V_1=526mL(0.516L)[/tex]

[tex]C_2=?[/tex]

[tex]V_2=?[/tex]

Now, we can calculate [tex]V_2[/tex] if we add the volumes, so:

[tex]0.516~L+~0.875~L=1.391~L[/tex]

So,  [tex]V_2=1.391~L[/tex]

We can plug the values in the equation:

[tex]3.82~M*0.516~L=C_2*1.391~L[/tex]

[tex]C_2=\frac{3.82~M*0.516~L}{1.391~L}[/tex]

[tex]C_2=1.42~M[/tex]

I hope it helps!

When 3.51 g of phosphorus was burned in chlorine, the product was a phosphorus chloride. Its vapor took 1.77 times as long to effuse as the same amount of CO2 under the same conditions of temperature and pressure. What is the molar mass of the phosphorus chloride

Answers

Answer:

molar mass of the phosphorus chloride = 138.06 g/mol

Explanation:

mass of phosphorus will be the same as mass of CO2, since it is stated that they are of equal amount.

mass = 3.51 g

lets assume that it took the CO2 1 sec to effuse, then the time taken by the phosphorus chloride will be 1.77 sec

From this we can say that

rate of effusion of CO2 = 3.51/1 = 3.51 g/s

rate of effusion of the phosphorus chloride = 3.51/1.77 = 1.98 g/s

From graham's equation of effusion,

[tex]\frac{Rc}{Rp}[/tex] = [tex]\sqrt{\frac{Mp\\}{Mc} }[/tex]

Rc = rate of effusion of CO2 = 3.51 g/s

Rp = rate of effusion of phosphorus chloride = 1.98 g/s

Mc = molar mass of CO2 = 44.01 g/mol

Mp = molar mass of the phosphorus chloride = ?

Imputing values into the equation, we have

[tex]\frac{3.51}{1.98}[/tex] = [tex]\sqrt{\frac{Mp\\}{44.01} }[/tex]

1.77 = [tex]\frac{\sqrt{Mp} }{6.64}[/tex]

11.75 = [tex]\sqrt{Mp}[/tex]

Mp = [tex]11.75^{2}[/tex]

Mp = molar mass of the phosphorus chloride = 138.06 g/mol

How many moles of hydrogen atoms are there in 120 g of C6H12O6

Answers

Answer: 8.0 moles

Explanation:

0.6661 moles×12 H≈8.0 moles

120 g of C₆H₁₂O₆ contain 8 moles of Hydrogen atoms

To obtain the answer to the question given above, we'll begin by calculating the mass of 1 mole of C₆H₁₂O₆

1 mole of C₆H₁₂O₆ = (12×6) + (12×1) + (16×6)

= 72 + 12 + 96

= 180 g

From the above,

We can see that 180 g of C₆H₁₂O₆ contains 12 moles of Hydrogen.

Finally, we shall determine the number of mole of Hydrogen atoms in 120 g of C₆H₁₂O₆. This can be obtained as follow:

180 g of C₆H₁₂O₆ contains 12 moles of Hydrogen.

Therefore,

120 g of C₆H₁₂O₆ will contain = [tex]\frac{120 * 12}{180}[/tex] = 8 moles of Hydrogen.

Thus, 120 g of C₆H₁₂O₆ contain 8 moles of Hydrogen atoms

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Acetic acid has a pKa of 4.74. Buffer A: 0.10 M HC2H3O2, 0.10 M NaC2H3O2 Buffer B: 0.30 M HC2H3O2, 0.30 M NaC2H3O2 Buffer C: 0.50 M HC2H3O2, 0.10 M NaC2H3O2 Which of the above buffers has the highest buffer capacity? [ Select ] Which of the above buffers has the lowest buffer capacity? [ Select ]

Answers

From the data provided;

Buffer B has the highest buffer capacityBuffer C has the lowest buffer capacityWhat is a buffer?

A buffer is a solution which resists changes to its pH when a small quantity of acid or base is added to it.

The buffer capacity is determined using the Henderson-Hasselbach equation:

pH = pKa + log ([CH3COO-]/[CH3COOH])

For Buffer A: pH = 4.74 + log(0.10/010) = 4.74

Buffer B: pH = 4.74 + log(0.30/030) = 4.74

For Buffer C: pH = 4.74 + log (0.10/0.50) = 4.04

Buffer A and Buffer B has same pH value as the pKa of acetic acid.

However, Buffer B has higher concentration of the components compared to buffer A, therefore, Buffer B has the highest buffer capacity.

The pH of buffer C is lower than pKa of acetic acid. Hence, buffer C has the lowest buffer capacity.

Learn more about buffers at: https://brainly.com/question/22390063

Select the correct answer from each drop-down menu.


Describe an oxidation-reduction reaction.


In an oxidation-reduction reaction, oxidation is what happens when a reactant___ . Reduction occurs when a reactant ___ in the reaction.

Answers

Explanation:

First off, we have to understand what we means by oxidation and reduction.

Oxidation is the loss of electrons during a reaction by a molecule, atom or ion. Oxidation occurs when the oxidation state of a molecule, atom or ion is increased.

Reduction on the other hand is the opposite of oxidation. It is the gain of electrons during a reaction by a molecule, atom or ion. Reduction occurs when the oxidation state of a molecule, atom or ion decreases.

Back to the question;

oxidation is what happens when a reactant loses electrons.

Reduction occurs when a reactant gains electrons in the reaction.

Which is a substance that could be found in air , water , or soil that is harmful to humans or animals?

Answers

Answer: Pollutant

Explanation:

Answer:

Explanation:

The substance that could be found in air, water or soil that is harmful to humans or animals is called pollutant.

PLEASE HELP I HAVE LIMITED TIME!!
How many particles are in 0.500 moles of
N.05?

Answers

Answer:

3.01×10²³ particles.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ particles. This implies that 1 mole of N2O5 also contains 6.02×10²³ particles.

Now, if 1 mole of N2O5 contains 6.02×10²³ particles ,

Then 0.5 mole of N2O5 will contain = 0.5 × 6.02×10²³ = 3.01×10²³ particles.

Therefore, 0.5 mole of N2O5 contains 3.01×10²³ particles.

Is electricity an essential property of matter ?

Answers

Answer: I think so.. (sorry if i am wrong)

Explanation:  Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity. A physical property is a characteristic of matter that is not associated with a change in its chemical composition.

An enzyme can catalyze a reaction with either of two substrates, S1 or S2. The Km for S1 was found to be 2.0 mM, and the Km, for S2 was found to be 20 mM. A student determined that the Vmax was the same for the two substrates. Unfortunately, he lost the page of his notebook and needed to know the value of Vmax. He carried out two reactions: one with 0.1 mM S1, the other with 0.1 mM S2. Unfortunately, he forgot to label which reaction tube contained which substrate. Determine the value of Vmax from the results he obtained:

Answers

Answer:

101

Explanation:

Provided that

[tex]S_1 = S_2 = same\ V_{max}[/tex]

And,

[tex]S_1\ k_M = 2.0mM\\S_2\ k_M = 20mM[/tex]

Now we expect the same

{S} (0.1mM)

This determines that [tex]S_1[/tex] generates a  higher rate of product formation as compared to the [tex]S_2[/tex]

So we can easily calculate the [tex]V_{max}[/tex] for either of [tex]S_1[/tex] or [tex]S_2[/tex] as we know that Tube 1 is [tex]S_2[/tex] and tube 2 is [tex]S_1[/tex]

As we know that

[tex]V_0 = V_{max}\ {S} / (K_M + {S})[/tex]

As the rates do not include any kind of units so we do not consider the units for [tex]V_{max}[/tex]

Now the calculation is

[tex]0.5 = V_{max} (0.1\ mM) / (20\ mM + 0.1\ mM)[/tex]

[tex]V_{max} = 0.5 (20.1\ mM) / 0.1\ mM[/tex]

= 100.5

101

If you are diving and playing in the waves near a beach and suddenly find that you are being

pulled out into the ocean at a relatively quick pace, what current are you most likely trapped in?

Answers

Answer: Wind current

Explanation:

The wind current is the main driving force along with the water currents when one reaches to the seashore. The wind currents above surface of water generates due to the evaporation of water.

The strong wind currents can pull any animal or human walking or running near the sea shore. These currents bring the animal or human towards the sea.

An electrochemical cell contains a standard hydrogen electrode and a cathode consisting of a metallic chromium electrode, Cr(s), in contact with a 1.00 M chromium solution, Cr3 (aq). The voltage produced by this cell was measured at 25oC. Which statements describe the results of this measurement, assuming the conditions are ideal? The cell voltage with the appropriate sign equals

I. the cell potential.
II. the electromotive force.
III. the standard cell potential.
IV. the standard reduction potential for Cr3+/Cr.

a. I only
b. l and Il
c. I, II and III
d. I, II, II, and IV
e. ll only

Answers

Answer:

c. I, II and III

Explanation:

The cell is as follows

Cr / Cr⁺²(1M) // H⁺ ( 1 M ) / H₂

Standard reduction potential of hydrogen cell is zero . Standard reduction potential is negative E for Cr⁺² / Cr(1M) half cell

Cell potential = Ecathode - Eanode

= 0 - ( - E)

= E

E is cell potential and also standard cell potential or emf of the cell .

Standard reduction potential that is for Cr3+/Cr.  is  - E .

Hence statement I , II , III are right . IV th statement is wrong because of sign

Option c is correct.

The branch of chemistry which deals with electricity is called electrochemistry.

The correct answer is C

The cell representation is as follows:-

[tex]Cr / Cr^{2+}(1M) // H^+ ( 1 M ) / H_2[/tex]

The standard reduction potential of hydrogen cells is zero. Standard reduction potential is negative E for Cr⁺² / Cr(1M) half cell because it will behave as a cathode.

The formula of the electric cell is as follows :-[tex]Cell potential = E_{cathode} - E_{anode[/tex]

After putting the value the cell potential will be:-

Cell potential = 0 - ( - E)

The cell potential will be= E

The standard reduction potential that is for [tex]Cr^{3+}/Cr = - E .[/tex]

Hence, the correct option is C that is I, II, and IV.

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What volume does 49.0g of KBr need to be dissolved in to make a 8.48% (w/v) solution?

Answers

Answer:

578mL is the volume you need to dissolve the mass of KBr

Explanation:

The percent by mass volume, % (w/v) is defined as the ratio between mass of solute in grams and volume of the solution in mililiters times 100. The formula is:

% (w/v) = mass solute (g) / volume solution (mL) × 100

As you want a solution 8.48% and the solute is 49.0g of KBr:

8.48% = 49.0 (g) / volume solution (mL) × 100

Volume solution = 578mL is the volume you need to dissolve the mass of KBr

Sulfurous acid is a diprotic acid with the following acid-ionization constants: Ka1 = 1.4x10−2, Ka2 = 6.5x10−8 If you have a 1.0 L buffer containing 0.252 M NaHSO3 and 0.139 M Na2SO3, what is the pH of the solution after addition of 50.0 mL of 1.00 M NaOH? Enter your answer numerically to 4 decimal places.

Answers

Answer:

pH = 7.1581

Explanation:

The equilibrium of NaHSO₃ with Na₂SO₃ is:

HSO₃⁻ ⇄ SO₃²⁻ + H⁺

Where K of equilibrium is the Ka2: 6.5x10⁻⁸

HSO₃⁺ reacts with NaOH, thus:

HSO₃⁻ + NaOH → SO₃²⁻ + H₂O + Na⁺

As the buffer is of 1.0L, initial moles of HSO₃⁻ and SO₃²⁻ are:

HSO₃⁻: 0.252 moles

SO₃²⁻: 0.139 moles

Based on the reaction of NaOH, moles added of NaOH are subtracting moles of HSO₃⁻ and producing SO₃²⁻. The moles added are:

0.0500L ₓ (1mol /L): 0.050 moles of NaOH.

Thus, final moles of both compounds are:

HSO₃⁻: 0.252 moles - 0.050 moles = 0.202 moles

SO₃²⁻: 0.139 moles + 0.050 moles = 0.189 moles

Using H-H equation for the HSO₃⁻ // SO₃²⁻ buffer:

pH = pka + log [SO₃²⁻] / [HSO₃⁻]

Where pKa is - log Ka = 7.187

Replacing:

pH = 7.187 + log [0.189] / [0.202]

pH = 7.1581

Consider two solutions: one formed by adding 10.0 grams of glucose (C6H12O6) to 1.0 L of water
and another formed by adding 10.0 grams of sucrose (C12H22O11) to 1.0 L of water. Which of the
two solutes produces the lower reduction in the vapor pressure of the solvent? Explain fully.

Answers

Answer:

Explanation:

Lowering of vapour pressure of solvent is proportional to number of moles   of solute dissolved in it per litre .

No of moles of glucose dissolved

= mass dissolved / molecular weight of glucose

= 10 / 180

= .055  moles.

No of moles of sucrose dissolved = 10 / 342

= .029 moles.

So reduction in vapour pressure will be lower of  solution dissolving sucrose . It is so because , no of moles of solute dissolved in it is low.  

Question 2
Bromine will react very fast (almost instantly) with which compound?
O 1-Pentene
Cyclohexane
Heptane
Benzene​

Answers

Answer:

1-Pentene

Explanation:

If we look at all the options listed, we will notice that the rate of reaction of bromine with each one differs significantly.

For 1-pentene, addition of bromine across the double bond is a relatively fast process. It is usually used as a test for unsaturation. Bromine water is easily decolorized by alkenes.

Cyclohexane, heptane are alkanes. They can only react with chlorine in the presence of sunlight. This is a substitution reaction. It does not occur easily. A certain quantum of light is required for the reaction to occur.

For benzene, bromine can only react with it by electrophilic substitution in which the benzene ring is retained. A Lewis acid is often required for the reaction to occur and it doesn't occur easily.

A student wishes to prepare 65.0 mL of 0.875 M HCl from 12.0 M HCl.

What volume of the
12.0 M HCl should he/she start with?

Answers

A student should start with 4.74 mL of the 12.0 M HCl to prepare 65.0 mL of 0.875 M HCl from 12.0 M HCl.

What is Molarity ?

Molarity (M) is the amount of a substance in a certain volume of solution.

Molarity is defined as the moles of a solute per liters of a solution.

Molarity is also known as the molar concentration of a solution.

Formula used for the given question ;

M₁V₁ = M₂V₂

M₁ = 12.0 M HCl.V₁ = ?M₂ = 0.875 M HCV₂ =  65.0 mL

M₁V₁ = M₂V₂

   V₁      =  0.875M x 65.0 mL / 12.0 M HCl.

            =  4.74 ml

Therefore , A student should start with 4.74 mL of the 12.0 M HCl to prepare 65.0 mL of 0.875 M HCl from 12.0 M HCl.

Learn more about Molarity here ;

https://brainly.in/question/11172828

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need help asappp please

Answers

Answer:

C.POLAR

Explanation:

According to research :

Methanol is a polar molecule:

The alcohol (-OH) group dominates the molecule making it definitely polar.

Oxygen in air can be removed using​

Answers

Answer:Fractional distillation of air. About 78 per cent of the air is nitrogen and 21 per cent is oxygen. These two gases can be separated by fractional distillation of liquid air.

Explanation:

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