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2 years ago

Why do we use the term formula unit for ionic compounds instead of the term molecule? Select all that apply.

Chemistry

1 answer:

VladimirAG [237]2 years ago

3 0

Answer:

A and C

Explanation:

The term formula unit is use to indicate Simple-Whole Ratios of ions in a compound. KEY WORD! RATIO!

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Acetylene (c2h2) undergoes combustion in excess oxygen to generate gaseous carbon dioxide and water. given δh°f[co2(g)] = –393.5

Brums [2.3K]

Answer is: 13181,7 kJ of energy is released when 10.5 moles of acetylene is burned.
Balanced chemical reaction: C₂H₂ + 5/2O₂ → 2CO₂ + H₂O.
ΔHrxn = sum of ΔHf (products of reaction) - sum of ΔHf (reactants).
Or ΔHrxn = ∑ΔHf (products of reaction) - ∑ΔHf (reactants).
ΔHrxn - enthalpy change of chemical reaction.
ΔHf - enthalpy of formation of reactants or products.
ΔHrxn = (2·(-393,5) + (-241,8)) - 226,6 · kJ/mol.
ΔHrxn = -1255,4 kJ/mol.
Make proportion: 1 mol (C₂H₂) : -1255,4 kJ = 10,5 mol(C₂H₂) : Q.
Q = 13181,7 kJ.

8 0

2 years ago

The Loch Ness monster is a sea serpent that is thought to live in a lake in

Dmitry_Shevchenko [17]

Answer:

Scotland

Explanation:

5 0

2 years ago

Read 2 more answers

Which of the following statements is not accurate? A. At constant temperature, halving the number of gas particles in a given sp

vlabodo [156]

I suggest you to go with C !! it is absolutely wrong sentence !!

increasing the gas particles will increase the collision between them so !!

your answer is C !!

4 0

3 years ago

Read 2 more answers

During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

8 0

3 years ago

What happens when you combine ammonia and formaldehyde?

bija089 [108]

Answer:

The mechanism for the formation of hexamethylenetetraamine predicts the formation of aminomethanol from the addition of ammonia to formaldehyde. This molecule subsequently undergoes unimolecular decomposition to form methanimine and water.

Explanation:

Brainliest please?

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2 years ago