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sergeinik [125]
3 years ago
10

1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.

Chemistry
1 answer:
Andrej [43]3 years ago
8 0

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

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5 0
3 years ago
Which of the following best describes how an ionic compound dissolves in water?
Nadusha1986 [10]

The statement that best describes how an ionic compound dissolves in water is as follows: it separates into individual molecules and is an electrolyte, which is option C.

<h3>What is an ionic compound?</h3>

Ionic compound is any compound is a chemical compound composed of ions (charged atoms) held together by electrostatic forces termed ionic bonding.

Ionic compounds are electrolytes i.e. a substance when, in solution or when molten, ionizes and conducts electricity.

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Therefore, the statement that best describes how an ionic compound dissolves in water is as follows: it separates into individual molecules and is an electrolyte.

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4 0
2 years ago
A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO
Akimi4 [234]

Answer: The empirical formula for the given compound is CH_2

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

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where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of CO_2=22.0g

We know that:

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For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, \frac{12}{44}\times 22.0=6g of carbon will be contained.

For calculating the mass of hydrogen:

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Mass of hydrogen = 1.0 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.5 moles.

For Carbon = \frac{0.5}{0.5}=1

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Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is C_{1}H_{2}=CH_2

4 0
3 years ago
In stomach, Hydrochloric acid kills micro-organisms in the food. Stomach juices begin to break down p to amino acids. Stomach ju
Butoxors [25]

\huge{ \underline{ \boxed{ \bf{ \green{Answer:}}}}}

Inside the stomach, Hydrochloric acid kills micro-organisms in the food. Stomach juices begin to break down <u>proteins</u> to amino acids.

✤ So, Fill the blank with proteins.

<h3><u>Explanation:-</u></h3>
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  • But this enzyme remains inactive and is activated by the Hydrochloric acid(HCl).
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<u>━━━━━━━━━━━━━━━━━━━━</u>

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Answer:

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