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3 years ago

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Which of the following is a synthesis reaction? 2Na + 2H 2O 2NaOH + H 2 2S + 3O 2 2SO 3 2H 2O 2 2H 2O + O 2 CaCl 2 + F 2 CaF 2 +

Cl 2

1 answer:

storchak [24]3 years ago

4 0

Answer:

2S + 3O 2 --->2SO 3 synthesis

Explanation:

2Na + 2H 2O -->2NaOH + H 2 single replacement

2S + 3O 2 --->2SO 3 synthesis

2H 2O 2 --->2H 2O + O 2 decomposition

CaCl 2 + F 2 ---> CaF 2 + Cl 2 single replacement

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A joule is a unit of _____. A. force B. work only C. energy only D. work or energy

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I would personally say C. Energy only...

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2 years ago

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150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac

Murljashka [212]

Answer:

$0.175\; \rm mol \cdot L^{-1}$ .

Explanation:

Magnesium chloride and silver nitrate reacts at a $2:1$ ratio:

$\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)$ .

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

$\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}$ .

The precipitate silver chloride $\rm AgCl$ is insoluble in water and barely ionizes. Hence, $\rm AgCl\!$ isn't rewritten as ions.

Net ionic equation:

$\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}$ .

Calculate the initial quantity of nitrate ions in the mixture.

$\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}$ .

Since nitrate ions $\rm {NO_3}^{-}$ do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

$n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol$ .

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be $(1/2)$ of the concentration in the original solution.

$\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}$ .

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2 years ago

Part APart complete A sample of sodium reacts completely with 0.568 kg of chlorine, forming 936 g of sodium chloride. What mass

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Answer: 368 grams of sodium reacted.

Explanation:

The balanced reaction is :

$2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)$

$\text{Moles of chlorine}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of chlorine}=\frac{0.568\times 1000g}{71g/mol}=8moles$

$\text{Moles of sodium chloride}=\frac{936g}{58.5g/mol}=16mol$

According to stoichiometry :

2 moles of $NaCl$ are formed from = 2 moles of $Na$

Thus 16 moles of $NaCl$ are formed from= $\frac{2}{2}\times 16=16moles$ of $Na$

Mass of $Na=moles\times {\text {Molar mass}}=16moles\times 23g/mol=368g$

Thus 368 grams of sodium reacted.

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2 years ago

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Answer:

the surface

Explanation:

n/a

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3 years ago

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Use the References to access important values if needed for this question There are 12 eggs in a dozen. If a farmer's chickens p

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Answer:

$3.4\times 10^3 eggs$ are produces are in a month.

Explanation:

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1 dozen = 12 eggs

Number of eggs in a month:

$284 dozens = 284\times 12 eggs =3,408 eggs\approx 3.4\times 10^3 eegs$

$3.4\times 10^3 eggs$ are produces are in a month.

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2 years ago