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ddd [48]
4 years ago
5

Hydrogen reacts with 0.771 g of carbon to form 0.90 g of a compound. What is the mass percent of hydrogen in the compound

Chemistry
2 answers:
morpeh [17]4 years ago
4 0
Answer: 14.3%

Explanation: In order to find the mass percent of hydrogen in this compound, you must determine how many grams of hydrogen you'd get in 100 g of compound. 

In your case, you know that an unknown mass of hydrogen reacts with 0.771 g of carbon to form 0.90 g of hydrocarbon, which is a compound that contains only carbon and hydrogen. 

Use the total mass of the hydrocarbon to determine how many grams of hydrogen reacted with the carbon.

Now, if 0.90 g of this compound contain 0.129 g of hydrogen, it follows that 100 g of this compound will contain.

So, if 100 g of this compound contain 14.33 g of hydrogen, it follows that the mass percent of hydrogen is 14.3%


<span><span><span><span><span><span><span><span><span /></span></span></span></span></span></span></span></span><span><span>
</span></span>
Dennis_Churaev [7]4 years ago
3 0
14.33% I hope this helps
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Answer:

0.48 V

Explanation:

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Zn(s) + Co^2+(aq) -------------> Zn^2+(aq) + Co(s) overall redox equation

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E°cell= E°cathode - E°anode

E°cell= -0.28-(-0.76)= 0.48 V

7 0
4 years ago
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How much of a sample remains after three half-lives have occurred? 1/16 of the original sample 1/9 of the original sample 1/8 of
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5 0
3 years ago
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A scientist measures the standard enthalpy change for the following reaction to be -17.2 kJ : Ca(OH)2(aq) 2 HCl(aq)CaCl2(s)
elixir [45]

Answer: \Delta H^{0}=-173.72 kJ/mol

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The standard enthalpy of formation of HCl is calculated as:

\Delta ^{0}=\Sigma H_{products}-\Sigma H_{reactants}

Ca(OH)_{2}_{(aq)}+2HCl_{(aq)} → CaCl_{2}_{(s)}+2H_{2}O_{(l)}

Standard Enthalpy of formation for the other compounds are:

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Water: \Delta H^{0}= -285.83 kJ/mol

Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.

Calculating:

-17.2=[-795.8+2(285.85)]-[-1002.82+2\Delta H]

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8 0
3 years ago
How many moles of HNO3 are present if 4.90×10−2 mol of Ba(OH)2 was needed to neutralize the acid solution?
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Answer:

0.098 moles

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Let y represent the number of moles present

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Hence, 0.049 moles of Ba(OH)2 contains y moles of OH- ions.

To get the y moles, we then do cross multiplication

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y mole = 2 * 0.049 / 1

y mole =  0.098 moles of OH- ions.

1 mole of OH- can neutralize 1 mole of H+

Therefore, 0.098 moles of HNO₃ are present.

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