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Aloiza [94]
2 years ago
14

a chemist dissolves 0.564 moles of manganese (IV) oxide (MnO2) in water, and adds enough water to make 0.510 L of solution. Calc

ulate the molarity of the solution.
Chemistry
1 answer:
ladessa [460]2 years ago
4 0

Answer:

The molarity of the solution is 1.1 \frac{moles}{liter}

Explanation:

Molarity is a measure of the concentration of that substance that is defined as the number of moles of solute divided by the volume of the solution.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

molarity=\frac{number of moles of solute}{volume}

Molarity is expressed in units \frac{moles}{liter}

In this case

  • number of moles of solute= 0.564 moles
  • volume= 0.510 L

Replacing:

molarity=\frac{0.564 moles}{0.510 L}

Solving:

molarity= 1.1 \frac{moles}{liter}

<u><em>The molarity of the solution is 1.1 </em></u>\frac{moles}{liter}<u><em></em></u>

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2.56 g of hydrogen reacts completely with 20.32 g of oxygen<br> to form X g of water. X = g
Brilliant_brown [7]

Answer:

Mass of water produced is 22.86 g.

Explanation:

Given data:

Mass of hydrogen = 2.56 g

Mass of oxygen = 20.32 g

Mass of water = ?

Solution:

Chemical equation:

2H₂ + O₂   →  2H₂O

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 20.32 g/ 32 g/mol

Number of moles = 0.635 mol

Number of moles of hydrogen:

Number of moles = mass/ molar mass

Number of moles = 2.56 g/ 2 g/mol

Number of moles = 1.28 mol

Now we will compare the moles of water with oxygen and hydrogen.

                    O₂            :            H₂O

                     1              :             2

                  0.635        ;            2×0.635 =  1.27

                   H₂             :              H₂O

                    2              :              2

                 1.28            :           1.28

The number of  moles of water produced by oxygen are less thus it will be limiting reactant.

Mass of water produced:

Mass = number of moles × molar mass

Mass = 1.27 × 18 g/mol

Mass = 22.86 g

 

4 0
3 years ago
Read 2 more answers
A 2.20 mol sample of NO 2 ( g ) is added to a 3.50 L vessel and heated to 500 K. N 2 O 4 ( g ) − ⇀ ↽ − 2 NO 2 ( g ) K c = 0.513
igor_vitrenko [27]

Answer:

[NO₂] = 0.434 M

[N₂O₄] = 0.0971 M

Explanation:

The equilibrum is:  N₂O₄(g)  ⇆  2NO₂ (g)

1 moles of nitrogen (IV) oxide is in equilibrium with 2 moles of nitrogen dioxide.

Initally we only have 2.20 moles of NO₂. So let's write the equilibrium again:

              2NO₂ (g)   ⇆   N₂O₄(g)      

Initially   2.20 mol              -

React          x                      x/2

X amount has reacted, and the half has been formed, according to stoichiometry.

Eq       (2.20-x) / 3.50L     (x/2)/ 3.50L

We divide by the volume because we need molar concentrations. Let's make the Kc's expression:

Kc = [N₂O₄] / [NO₂]²

0.513 = ((x/2)/ 3.50L) /  [(2.20-x) / 3.50L]

0.513 = ((x/2)/ 3.50L) / [(2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / [2.20-x)² / 3.50L²]

0.513 = ((x/2)/ 3.50L) / (4.84 - 4.40x + x²) / 12.25)

0.513 / 12.25 (4.84 - 4.40x + x²) = x/2 / 3.50

0.203 - 0.184x + 0.0419x² = x/2 / 3.50

3.50(0.203 - 0.184x + 0.0419x²) = x/2

7 (0.203 - 0.184x + 0.0419x²) - x = 0

1.421 - 2.288x + 0.2933x² = 0  → Quadratic formula

a = 0.2933 ;  b = -2.288 ; c = 1.421

(-b +- √(b²-4ac)) / (2a)

x₁ = 7.12

x₂ = 0.68 → We consider this value, so we can have a (+) concentration.

Concentrations in the equilibrium are:

[NO₂] = (2.20-0.68) / 3.50 = 0.434 M

[N₂O₄] = (0.68/2) / 3.50  = 0.0971 M

8 0
3 years ago
What is the MOLAR heat of combustion of methane(CH₄) if 64.00g of methane are burned to heat 75.0 ml of water from 25.00°C to 95
melamori03 [73]

Answer:

-5.51 kJ/mol

Explanation:

Step 1: Calculate the heat required to heat the water.

We use the following expression.

Q = c \times m \times \Delta T

where,

  • c: specific heat capacity
  • m: mass
  • ΔT: change in the temperature

The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.

Q = 4.184J/g.\°C \times 75.0g \times (95.00\°C - 25.00\°C) = 2.20 \times 10^{3} J = 2.20 kJ

Step 2: Calculate the heat released by the methane

According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero

Qc + Qw = 0

Qc = -Qw = -22.0 kJ

Step 3: Calculate the molar heat of combustion of methane.

The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.

\frac{-22.0kJ}{64.00g} \times \frac{16.04g}{mol} = -5.51 kJ/mol

8 0
3 years ago
For all of the following questions 20.00 mL of 0.200 M HBr is titrated with 0.200 M KOH.
HACTEHA [7]

Answer :

The concentration of H^+ before any titrant added to our starting material is 0.200 M.

The pH based on this H^+ ion concentration is 0.698

Explanation :

First we have to calculate the concentration of H^+ before any titrant is added to our starting material.

As we are given:

Concentration of HBr = 0.200 M

As we know that the HBr is a strong acid that dissociates complete to give hydrogen ion H^+ and bromide ion Br^-.

As, 1 M of HBr dissociates to give 1 M of H^+

So, 0.200 M of HBr dissociates to give 0.200 M of H^+

Thus, the concentration of H^+ before any titrant added to our starting material is 0.200 M.

Now we have to calculate the pH based on this H^+ ion concentration.

pH : It is defined as the negative logarithm of hydrogen ion concentration.

pH=-\log [H^+]

pH=-\log (0.200)

pH=0.698

Thus, the pH based on this H^+ ion concentration is 0.698

3 0
3 years ago
A 1.000-g sample of lead shot reacted with oxygen to give 1.077 g of product. Calculate the empirical formula of the lead oxide
Evgesh-ka [11]
It mean it consisted of 1 g of lead and 0.077 g of O2.
divide these numbers by molar mas.
1/82=0.012 Pb /0.004 = 3
0.077/16= 0.004 O /0.004 =1
Pb3O

3 0
3 years ago
Read 2 more answers
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