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3 years ago

Chlorine atom

Chemistry

1 answer:

irina [24]3 years ago

8 0

Answer:

C. Decomposition of potassium permanganate(Heating)

Explanation:

The equation of the reaction is given as;

2KMnO4 -> K2MnO4 + MnO2(s) + O2(g)

Reactant = 2KMnO4

Products = K2MnO4 + MnO2(s) + O2(g)

A. Oxygen turned in to carbon dioxide

Incorrect option - Oxygen is not the reactant

B. Sulphate and lithium boiled

Incorrect option - Sulphate and lithium are not part of this reaction

C. Decomposition of potassium permanganate(Heating)

Correct option - potassium permanganate decomposed to form K2MnO4 + MnO2(s) + O2(g)

D. None of the above

Incorrect option

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Consider the reaction between acetylene, c2h2, and oxygen in a welding torch: 2c2h2(g) + 5o2(g) → 4co2(g) + 2h2o(g) if 5.4 moles

fredd [130]

2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
from the reaction 2 mol 4 mol
from the problem 5.4 mol 10.8 mol

M(CO2) = 12.0 +2*16.0 = 44.0 g/mol
10.8 mol CO2 * 44.0 g CO2/1 mol CO2 = 475.2 g CO2 ≈480 = 4.8 * 10² g
Answer is C. 4.8*10² g.

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3 years ago

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Explain how humans acquire the energy to function.Explain the process of cellular respiration and how we obtain this energy.

Naddika [18.5K]

Answer:

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Explanation:

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3 years ago

Calculate the molar concentration of the Cl⁻ ions in 0.65 M CaCl2(aq), assuming that the dissolved substance dissociates complet

alexandr1967 [171]

The question here is solved using basic chemistry. CaCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of CaCl2 that dissolves.
CaCl2(s) --> Ca+(aq) + 2 Cl⁻(aq)
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4 years ago

Vanyuwa [196]

Answer:

If the volume of a gas increased from 2 to 6 L while the temperature was held constant, the pressure of the gas decreased by a factor of 3.

Explanation:

Boyle's law that says "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

P * V = k

To obtain the proportionality factor k you must make the quotient:

$k=\frac{V2}{V1} =\frac{6 L}{2 L}$

k= 3

This means that if the volume of a gas increased from 2 to 6 L while the temperature was held constant, the pressure of the gas decreased by a factor of 3.

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3 years ago

7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.

beks73 [17]

Answer:

$\Delta H_{f,C_3H_4}=276.8kJ/mol$

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

$\Delta H_{rxn} =- m_wC_w\Delta T$

We plug in the mass of water, temperature change and specific heat to obtain:

$\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ$

Now, this enthalpy of reaction corresponds to the combustion of propyne:

$C_3H_4+4O_2\rightarrow 3CO_2+2H_2O$

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

$\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}$

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

$\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol$

Now, we solve for the enthalpy of formation of C3H4 as shown below:

$\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}$

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

$\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol$

Best regards!

7 0

3 years ago