1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ad-work [718]
4 years ago
6

What is the molarity of a 5.00x10^2 mL solution containing 21.1 g of potassium bromide (KBr)? The molar mass of KBr is 119.0 g/m

ol.
a)3.55x10−4M
b)42.2M
c) 0.355 M
d)8.87∗10−2
Chemistry
2 answers:
3241004551 [841]4 years ago
8 0
Calculate the number of moles (n).
                             n = (21.1 g) x (1 mol / 119 g) = 211/1190 moles
Divide this number of moles by the volume of the solution,
             M = (21/1190 moles) / (500 mL x 1L/1000mL) = 0.355 M
The answer is letter C. 
Dominik [7]4 years ago
5 0
Volume of solution in liters:

5.00 x 10² mL / 1000 => 0.5 L 

number of moles:

mass of solute / molar mass

21.1 / 119.0 => 0.1773 moles

Molarity = number of moles / volume 

M = 0.1773 / 0.5

M = 0.355 mol/L

Answer C

<span>hope this helps!</span>
You might be interested in
5. The heat of fusion of ice is 333.5J/g. The entropy change for the water when freezing 5.0 g of water at 0°C
lara [203]

Answer:

(e) -6.1 J/K

Explanation:

Step 1: Given data

  • Heat of fusion of ice (ΔH°fus): 333.5 J/g
  • Mass of water (m): 5.0 g

Step 2: Calculate the heat (Qfreezing) required to freeze 5.0 g of water

We will use the following expression.

Qfreezing = -ΔH°fus × m

Qfreezing = -333.5 J/g × 5.0 g = -1.7 × 10³ J

Step 3: Calculate the entropy change (ΔS°) at 0 °C (273.15 K) and 1 atm

We will use the following expression.

ΔS° = Qfreezing/T

ΔS° = -1.7 × 10³ J/273.15 K = -6.1 J/K

7 0
3 years ago
Calculate the standard molar enthalpy for the complete combustion of liquid ethanol (c2h5oh) using the standard enthalpies of fo
olga2289 [7]
Answer : To calculate the standard molar enthalpy of complete combustion of liquid ethanol can be done by using the formula given below;

C_{2}  H_{5}OH + 3O_{2} ----\ \textgreater \   2CO_{2}  + 3 H_{2}O

Now, the enthalpies of the elements in their standard state is zero always. 

So the H_{reaction} =  H_{products}  - H_{reactants}

Here we can avoid the heat of enthalpy of oxygen,carbon dioxide and water as they exists in their standard states.

So the H_{reaction} = H_{ C_{2}H_{5}OH}
6 0
3 years ago
Read 2 more answers
For the following reaction, 5.20 grams of propane (C3H8) are allowed to react with 22.5 grams of oxygen gas. propane (C3H8) (g)
Sliva [168]

Answer:

15.58g of CO₂ is the maximum amount that can be produced

Explanation:

The propane reacts with oxygen as follows:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

<em>Where 1 mole of propane reacts with 5 moles of oxygen</em>

To solve this question we need to find the moles of propane and oxygen to find limiting reactant using the chemical reaction:

<em>Moles propane -Molar mass: 44.1g/mol-:</em>

5.20g * (1mol / 44.1g) = 0.118 moles

<em>Moles oxygen -Molar mass: 32g/mol-:</em>

22.5g O₂ * (1mol / 32g) = 0.703 moles

For a complete reaction of 0.703 moles of oxygen are:

0.703 moles O₂ * (1mol C₃H₈ / 5mol O₂) = 0.141 moles of propane are necessaries. As there are just 0.118 moles of propane, <em>propane is limiting reactant.</em>

The moles of carbon dioxide that are produced are:

0.118 moles C₃H₈ * (3 moles CO₂ / 1 mol C₃H₈) = 0.354 moles CO₂

The maximum mass that can be produced is -Molar mass CO₂: 44.01g/mol-:

0.354 moles CO₂ * (44.01g / mol) =

15.58g of CO₂ is the maximum amount that can be produced

4 0
3 years ago
A(n)_______is a balanced chemical equation that includes the
Katyanochek1 [597]

Answer:

thermochemical equation

Explanation:

A(n)__thermochemical equation_____is a balanced chemical equation that includes the physical states of all reactants and products, and the energy change that accompanies the reaction.

7 0
3 years ago
The gas OF2 can be produced from the electrolysis of an aqueous solution of KF, as shown in the equation below.
Readme [11.4K]

Answer:

A) 6.48 g of OF₂ at the anode.

Explanation:

The gas OF₂ can be obtained through the oxidation of F⁻ (inverse reaction of the reduction presented). The standard potential of the oxidation is the opposite of the standard potential of the reduction.

H₂O(l) + 2 F⁻(aq) → OF₂(g) + 2 H⁺(aq) + 4 e⁻    E° = -2.15 V

Oxidation takes place in the anode.

We can establish the following relations:

  • 1 Faraday is the charge corresponding to 1 mole of e⁻.
  • 1 mole of OF₂ is produced when 4 moles of e⁻ circulate.
  • The molar mass of OF₂ is 54.0 g/mol.

The mass of OF₂ produced when 0.480 F pass through an aqueous KF solution is:

0.480F.\frac{1mole^{-} }{1F} .\frac{1molOF_{2}}{4mole^{-} } .\frac{54.0gOF_{2}}{1molOF_{2}} =6.48gOF_{2}

7 0
3 years ago
Other questions:
  • If a long steady rain is expected which clouds should be expected
    6·1 answer
  • When a molecule of nad+ gains a hydrogen atom, the molecule becomes reduced.
    14·1 answer
  • How is heavy water different from normal water
    13·1 answer
  • Here are approximately 15 milliliters (mL) in 1 tablespoon (tbsp).
    5·2 answers
  • Which group of people was President Jackson
    8·2 answers
  • Select all the correct answers.
    7·1 answer
  • Before a bond breaks in a chemical reaction, what happens
    9·1 answer
  • 5.6 tons of pig ovaries were required to extract 12.0 mg of estrone. Assuming complete isolation of the hormones was achieved ca
    14·1 answer
  • A paragraph on information about density.
    7·1 answer
  • What is NOT true about white light?
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!