PLEASE ANSWER

A 1.2516 gram sample of a mixture of caco3 and na2so4 was analyzed by dissolving the sample and completely precipitating the ca

Dennis_Churaev [7]

Answer:

0.009725 moles of H2C2O4

0.009725 moles CaCO3

Mass percentage = 77.77%

Explanation:

Step 1: The balanced equation

2MnO4- +5C2H2O4+6H+ →2Mn2+ +10CO2+8H2O

We can see that for 2 moles of Mno4- consumed , there is 5 moles of C2H2O4 needed and 6 moles H+ to produce 2 moles Mn2+, 10 moles of CO2 and 8 moles of H2O

Step 2: Calculate moles of MnO4-

Molarity = Moles/volume

Moles of Mno4- = Molarity of MnO4- * Volume of Mno4-

Moles of Mno4- = 0.1092M * 35.62 *10^-3 L

Moles of MnO4- = 0.00389 moles

Step 3: Calculate moles of H2C2O4

Since there is needed 5 moles of C2H2O4 to consume 2 moles of MnO4-

then for 0.00389 moles of MnO4-, there is 5/2 *0.00389 = 0.009725 moles of H2C2O4

Step 4: Calculate moles of CaCO3

moles of H2C2O4 = moles CaCO3, therefore, 0.009725 moles H2C2O4 = 0.009725 moles CaCO3

Step 5: Calculate mass of CaCO3

Molar mass of CaCO3 = 100.09 g/mole

Mass of CaCO3 = moles of CaCO3 * Molar mass of CaCO3

Mass of CaCO3 = 0.009725 moles * 100.09 g/mole = 0.9734 g

Step 6: Calculate percentage by weight of CaCO3

Mass of CaCO3 = 0.9734g

Mass of original sample = 1.2516g

Mass percentage = 0.9734/1.2516 *100% = 77.77%

6 0

3 years ago

A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution

Dmitrij [34]

Answer:

The change in internal energy is - 1.19 kJ

Explanation:

Step 1: Data given

Heat released = 3.5 kJ

Volume calorimeter = 0.200 L

Heat release results in a 7.32 °C

Temperature rise for the next experiment = 2.49 °C

Step 2: Calculate Ccalorimeter

Qcal = ccal * ΔT ⇒ 3.50 kJ = Ccal *7.32 °C

Ccal = 3.50 kJ /7.32 °C = 0.478 kJ/°C

Step 3: Calculate energy released

Qcal = 0.478 kJ/°C *2.49 °C = 1.19 kJ

Step 4: Calculate change in internal energy

ΔU = Q + W W = 0 (no expansion)

Qreac = -Qcal = - 1.19 kJ

ΔU = - 1.19 kJ

The change in internal energy is - 1.19 kJ

4 0

3 years ago

How many moles are contained in 2.3 liters of a 1.2M solution?

elena-s [515]

Answer:

$\boxed {\boxed {\sf 2.76 \ mol}}$

Explanation:

Molarity is found by dividing the moles of solute by liters of solution.

$molarity = \frac {moles}{liters}$

We know the molarity is 1.2 M (mol\liter) and there are 2.3 liters of solution. Substitute the known values into the formula.

$1.2 \ mol/liter= \frac {x}{2.3 \ liters}$

Since we are solving for x, we must isolate the variable. It is being divided by 2.3 and the inverse of division is multiplication. Multiply both sides by 2.3 liters.

$2.3 \ liters *1.2 \ mol/liter= \frac {x}{2.3 \ liters}* 2.3 \ liters\\2.3 *1.2 \ mol= x\\2.76 \ mol =x$

In a solution with a molarity of 1.2 and 2.3 liters of solution, there are 2.76 moles.

3 0

3 years ago

There are 7 named classes of hazardous materials. O True O False

Anika [276]

False

there is actually 9

7 0

2 years ago

Determine the volume of water to be added to the nitric acid solution at a concentration of 8.61 mol / L to prepare 500 mL of th

Alex_Xolod [135]

Answer:

398 mL

Explanation:

Using the equation for molarity,

C₁V₁ = C₂V₂ where C₁ = concentration before adding water = 8.61 mol/L and V₁ = volume before adding water, C₂ = concentration after adding water = 1.75 mol/L and V₂ = volume after adding water = 500 mL = 0.5 L

V₂ = V₁ + V' where V' = volume of water added.

So, From C₁V₁ = C₂V₂

V₁ = C₂V₂/C₁

= 1.75 mol/L × 0.5 L ÷ 8.61 mol/L

= 0.875 mol/8.61 mol/L

= 0.102 L

So, V₂ = V₁ + V'

0.5 L = 0.102 L + V'

V' = 0.5 L - 0.102 L

= 0.398 L

= 398 mL

So, we need to add 398 mL of water to the nitric solution.

6 0

3 years ago