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3 years ago

5

When reacting 45g of magnesium with oxygen, 50.0g of magnesium oxide is produced. Calculate the theoretical yield and percent yi

eld.

1 answer:

Veseljchak [2.6K]3 years ago

5 0

Hope you are able to get it :)

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10. What is the Kinetic Energy of a 100 kg object that is moving with a speed of 13.5 m/s?

marissa [1.9K]

Answer:

$100 \times 13.5 = ikw n bahaa$

IKAW na mag tayms

7 0

3 years ago

Hydrogen peroxide decomposes to water and oxygen at constant pressure by the following reaction: 2H2O2(l) → 2H2O(l) + O2(g) ΔH =

Mashcka [7]

<u>Answer:</u> The amount of heat released is -7.203 kJ

<u>Explanation:</u>

The given chemical equation follows:

$2H_2O_2(l)\rightarrow 2H_2O(l )+O_2(g);\Delta H=-196kJ$

To calculate the enthalpy change for 1 mole of the hydrogen peroxide, we use unitary method:

When 2 moles of hydrogen peroxide is reacted, the enthalpy of the reaction is -196 kJ

So, when 1 mole of hydrogen peroxide will react, the enthalpy of the reaction will be $\frac{-196}{2}\times 1=-98kJ$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of hydrogen peroxide = 2.50 g

Molar mass of hydrogen peroxide = 34 g/mol

Putting values in above equation, we get:

$\text{Moles of hydrogen peroxide}=\frac{2.50g}{34g/mol}=0.0735mol$

To calculate the heat of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

$q$ = amount of heat released

n = number of moles = 0.0735 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -98 kJ/mol

Putting values in above equation, we get:

$-98kJ/mol=\frac{q}{0.0735mol}\\\\q=(-98kJ/mol\times 0.0735mol)=-7.203kJ$

Hence, the amount of heat released is -7.203 kJ

8 0

3 years ago

Can someone pls tell me the answer to these 2 fill in the blanks I will surely mark him as the brain list answer

Elena-2011 [213]

Question 22: conclusion
Question 23: analyze

7 0

2 years ago

Considering the volumes and the concentrations used in mixture 1, what percentage of the moles of H2O2 present have been consume

Kisachek [45]

Answer:

1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2

Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.

Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.

The amount of I ions that is oxidized by H2O2 and the iodine, I2 that is produced, is consumed by the S2O32- ions and are transformed into I ions. The amount of I ions that are consumed by H2O2 is equal to the amount of ion that is produced by S2O32-.

Explanation:

the total volume of the mixture is equal to:

Vmix = 75 + 30 + 25 + 5 + 5 + 10 = 150 mL

the moles of each species in the mix equals:

Na2S2O3 = 0.05 * 0.005 = 0.00025 moles

KI = 0.05 * 0.025 = 0.00125 moles

H2O2 = 1.02 * 0.01 = 0.0102 moles

the following equation shows the reaction between I2 and S2O32:

I2 + S2O32 = 2I- + S4O62-

The same way:

2I- + 2H+ + H2O2 = I2 + 2H2O

1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2

Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.

Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.

The amount of I ions that is oxidized by H2O2 and the iodine, I2 that is produced, is consumed by the S2O32- ions and are transformed into I ions. The amount of I ions that are consumed by H2O2 is equal to the amount of ion that is produced by S2O32-.

3 0

3 years ago

Sodium carbonate is often used in certain candies to create fizzy bubbles when solid weak acids in the candy are dissolved with

erastovalidia [21]

Answer is: in chemical reaction with with weak acid, sodium carbonate produce carbonic acid (H₂CO₃), which decomposes into carbon(IV) oxide (fizzy bubbles) and water.

Balanced chemical reaction:

Na₂CO₃ + 2HClO → CO₂(g) + H₂O + 2NaClO.

Chemical decomposition is the separation of a carbonic acid into water and carbon dioxide: H₂CO₃ → CO₂ + H₂O.

6 0

3 years ago