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2 years ago

Which compound will conduct electricity when it is dissolved in water? CH4 CuSO4 C6H6 C6H12O6

2 answers:

6 0

Only CuSO4 conducts electricity in water because it is an ionic compound. The conduction of electricity is due to the ions in the ionic compound. When copper sulphate is dissolved in water, the ions dissociate and conduct electricity.
The others are do not liberate electricity when dissolved in water because ions are not liberated when electricity is passed through their solutions.

mestny [16]2 years ago

4 0

I would say CuSO4 or Copper Sulfate, as option 1 is Methane and would create a fire so heat, and the last one is Sugar which doesn't conduct electricity. And C6H6 I believe is not soluble in water.

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When the temperature of a gas in a rigid container decreases, the particles of the gas move slower and experience fewer collisio

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3 years ago

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Why must there be two chlorine atoms for every one calcium atom

Nady [450]

Answer:

There must be two Chlorine atoms for every one Calcium atom in order to fulfill Chlorine's octet rule and pair Calcium's unpaired electrons.

Explanation:

Calcium has two unpaired electrons in its Lewis dot structure, while Chlorine has one unpaired electron.

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Chlorine has seven valence electrons, so once it shares electrons with Calcium, the octet rule is accomplished, and no more pairs can be made.

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3 years ago

For each of the following ionic bonding examples, show the transfer of electrons, the ions resulting with the charges and the fi

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2 years ago

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

Andrej [43]

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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2 years ago

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