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2 years ago

What is the total mass of products formed when 50. 0 grams of CH4 is burned with excess oxygen? CH4 + 2 O2 → CO2 + 2 H2O

Chemistry

1 answer:

Alborosie2 years ago

7 0

Since 16 gm CH4 produce 36 gm H2O

Hence 2.5 gmCH4 produce 36×2.5/16 gm H2O

= 5.265 gm of H2O

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Andrews [41]

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6 0

3 years ago

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40. Write a balanced equation and a net ionic equation for combining AgNO3(aq) and Na2CO3(aq)

ki77a [65]

2AgNO3 (aq) + Na2CO3 (aq) --> Ag2CO3 (s) + 2NaNO3 (aq)

2Ag(+) (aq) + CO3(2-) (aq) --> Ag2CO3 (s)
hope it helps

3 0

4 years ago

2. In Experiment SOL, you investigated the solubility of oxalic acid. Sodium oxalate, Na2C2O¬4, is the sodium salt of this acid.

AnnZ [28]

Answer:

Sodium oxalate is a basic salt. In water it can be dissolved and dissociated.

The oxalic acid in water has two dissociations.

Explanation:

Na2C2O4 ---> 2Na+ + C2O4-2

Sodium oxalate is the conjugate base of a weak acid. In water this salt, dissociates completely giving rise to the sodium and oxalate ions. As Na+ comes from a strong base, in water it does not produce hydrolysis while oxalate does react in water, because it takes a proton from it and it generates a basic hydrolysis releasing OH-.

C2O4-2 + H2O ⇄ HC2O4- + OH-

In water the salt is basic. The pH of an aqueous solution of this salt is basic, since OH- is generated.

The HC2O4- has a second hydrolisis, it takes another proton from water to form oxalic acid.

HC2O4- + H2O ⇄ H2C2O4 + OH-

The oxalic acid acts as a weak acid, it can release 2 protons to water, to make oxalate (its conjugate base).

H2C2O4 + H2O ⇄ H3O+ + HC2O4-

HC2O4- + H2O ⇄ H3O+ C2O4-2

The HC2O4- acts as an ampholyte since it accepts and delivers protons simultaneously.

6 0

4 years ago

A word of equationn for the reacyion between lithium and water

Elenna [48]

Answer:

Equation: - 2Li + H2O = Li2O + H2 Uncoated lithium metal reacts with water to form a colorless lithium hydroxide solution and hydrogen gas.

Explanation:

4 0

3 years ago

the heat of fusion of acetone is 5.7 kJ/mol. Calculate to two significant figures the entropy change when 6.3 mol of acetone mel

shtirl [24]

<u>Answer:</u> The entropy change of the process is $2.0\times 10^2J/K$

<u>Explanation:</u>

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=n\times \frac{\Delta H_{f}}{T}$

where,

$\Delta S$ = Entropy change

n = moles of acetone = 6.3 moles

$\Delta H_{f}$ = enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol (Conversion factor: 1 kJ = 1000 J)

T = temperature of the system = $-94.7^oC=[273-94.7]=178.3K$

Putting values in above equation, we get:

$\Delta S=\frac{6.3mol\times 5700J/mol}{178.3K}\\\\\Delta S=201.4J/K=2.0\times 10^2J/K$

Hence, the entropy change of the process is $2.0\times 10^2J/K$

4 0

4 years ago