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3 years ago

Calculate the molarity of the solutions described below.

Chemistry

1 answer:

maria [59]3 years ago

7 0

Answer:

Explanation:

A. Na = 23 g/m Cl = 35.5 g/m NaCl = 58.5 g/m

10g / 58.5 g/m = .17 mole

.17 mole/ 2.0 L = 0.086 m/L = 0.086 Molar

B. 342 g/m 52.5/342 =0.15 mole 0.15 mole/1.0 L = 0.15 Molar

C. Al2(SO4)3== 342 g/m 120/342 = 0.35 mole 0.35 m/10.L 0.035 Molar

D caffeine = 194 g/m 1.75/ 194 = 0.009 mole 0.009/0.100 L 0.09 Molar

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One of relatively few reactions that takes place directly between two solids at room temperature is In this equation, the in Ba(

yuradex [85]

Answer:

3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate.

Explanation:

$Ba(OH)_2.8H_2O(s)+NH_4SCN(s)\rightarrow Ba(SCN)_2(s)+8H_2O(l)+NH_3(g)$

The balance chemical equation is :

$Ba(OH)_2.8H_2O(s)+2NH_4SCN(s)\rightarrow Ba(SCN)_2(s)+10H_2O(l)+2NH_3(g)$

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Moles of barium hydroxide octahydrate = $\frac{6.5 g}{315 g/mol}=0.020635 mol$

According to reaction, 2 moles of ammonium thiocyanate reacts with1 mole of barium hydroxide octahydrate. The 0.020635 moles of barium hydroxide octahydrate will react with:

$\frac{2}{1}\times 0.020635 mol=0.04127 mol$

Mass of 0.04127 moles of ammonium thiocyanate;

$0.04127 mol\times 76 g/mol=3.136 g\approx 3.14 g$

3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate

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3 years ago

Name the type of chemical reaction : CaO(s) + H2O(l) -----------Ca(OH)2(aq)

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It is called Synthesis reaction

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3 years ago

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2 years ago

4. Arrange the following substances in order of increasing boiling point: CH3CH2OH, HOCH2CH2OH, CH3CH2Cl , and ClCH2CH2OH A. CH3

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Answer: Option (B) is the correct answer.

Explanation:

A molecules that will have more number of hydrogen bonding will has highest boiling point because to break the hydrogen bonds high heat needs to be provided.

So, in the molecule $HOCH_{2}CH_{2}OH$ there is presence of two alcoholic groups. Hence, it will have strongest hydrogen bonding as compared to the rest of molecules.

In the molecule $ClCH_{2}CH_{2}OH$ there will be hydrogen bonding and dipole-dipole interactions. Hence, it boiling point will be slightly less than $HOCH_{2}CH_{2}OH$ .

In the molecule $CH_{3}CH_{2}OH$ , there will be only hydrogen bonding. Hence, its boiling point is less than $ClCH_{2}CH_{2}OH$ .

In the molecule $CH_{3}CH_{2}Cl$ , there is no hydrogen bonding but there will be only dipole-dipole interactions. Hence, its boiling point will be the least.

Therefore, we can conclude that increasing order of boiling point will be as follows.

$CH_{3}CH_{2}Cl$ < $CH_{3}CH_{2}OH$ < $ClCH_{2}CH_{2}OH$ < $HOCH_{2}CH_{2}OH$

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3 years ago