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Harlamova29_29 [7]
3 years ago
6

Which substance has the lowest boiling point?

Chemistry
2 answers:
Orlov [11]3 years ago
8 0

Answer:

Helium

Explanation:

tia_tia [17]3 years ago
4 0

Answer:

If you actually talking about substance like for example Salt - Sugar

i think Salt boiling point would be slowest

cause i think where your question came from

but if your talking about elememnt it would be Hellium

: ) hope it helped out

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A company is testing drinking water and wants to ensure that ca content is below 155 ppm. What is the maximum amount of Ca that
BabaBlast [244]

Answer:

138 mg

Explanation:

A company is testing drinking water and wants to ensure that Ca content is below 155 ppm (= 155 mg/kg), that is, <em>155 milligrams of calcium per kilogram of drinking water</em>. We need to find the maximum amount of calcium in 890 g of drinking water.

Step 1: Convert the mass of drinking water to kilograms.

We will use the relation 1 kg = 1000 g.

890g \times \frac{1kg}{1000g} =0.890kg

Step 2: Calculate the maximum amount of calcium in 0.890 kg of drinking water

0.890gH_2O \times \frac{155mgCa}{1kgH_2O} = 138mgCa

7 0
3 years ago
Jenny wants to test the electrical conductivity of two substances dissolved in water. She is preparing the containers for the ex
katen-ka-za [31]

Answer:

Volume of the solutions

This is the most important factor for her to control.

3 0
3 years ago
Which one is unsaturated hydrocarbon?
Fittoniya [83]

Answer:

The correct answer is - D C2H4.

Explanation:

Saturated hydrocarbons are hydrocarbons with single covalent C-C bonds. They are known as alkanes. The general formula for these hydrocarbons is  CnH2n+2

Unsaturated hydrocarbons the hydrocarbons with double or triple covalent C-C bonds. They are known as alkenes and alkynes respectively. The general formula for these hydrocarbons is  CnH2n and CnHn-2

For the given options:

Option D: C2H4, is the simplest alkene with a double bond so it is an unsaturated hydrocarbon.

6 0
3 years ago
For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox
ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: m_{N2}=10.682 g N_2

2) Limiting reagent: NO

3) Ammount of excess reagent: m_{N2}=4.274 g

Explanation:

<u>The reaction </u>

2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)

Moles of nitrogen monoxide

Molecular weight: M_{NO}=30 g/mol

n_{NO}=\frac{m_{NO}}{M_{NO}}

n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol

Moles of hydrogen

Molecular weight: M_{H2}=2 g/mol

n_{H2}=\frac{m_{H2}}{M_{H2}}

n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted

m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}

m_{N2}=10.682 g N_2

2) <u>Limiting reagent</u>: NO

3) <u>Ammount of excess reagent</u>:

m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}

m_{N2}=4.274 g

8 0
3 years ago
Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv
Triss [41]

Answer:

1. 3.57\times 10^{-4}was the K_a value calculated by the student.

2. 5.93\times 10^{-4}was the K_b of ethylamine value calculated by the student.

Explanation:

1.

The pH value of Aspirin solution = 2.62

pH=-\log[H^+]

[H^+]=10^{-2.62}=0.00240 M

Moles of s asprin = \frac{2.00 g}{180 g/mol}=0.01111 mol

Volume of the solution = 0.600 L

The initial concentration of Aspirin  = c = \frac{0.01111 mol}{0.600 L}=0.0185 M

HAs\rightleftharpoons As^-+H^+

initially

c       0    0

At equilibrium

(c-x)      x   x

The expression of dissociation constant :

K_a=\frac{[As^-][H^+]}{[HAs]}:

K_a=\frac{x\times x }{(c-x)}

=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}

K_a=3.57\times 10^{-4}

3.57\times 10^{-4}was the K_a value calculated by the student.

2.

The pH value of ethylamine = 11.87

pH+pOH=14

pOH=14-11.87=2.13

pOH=-\log[OH^-]

[OH^-]=10^{-2.13}=0.00741 M

The initial concentration of ethylamine = c = 0.100 M

C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-

initially

c                    0    0

At equilibrium

(c-x)                x   x

The expression of dissociation constant :

K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}:

K_b=\frac{x\times x}{(c-x)}

=\frac{0.00741\times 0.00741}{(0.100-0.00741)}

K_b=5.93\times 10^{-4}

5.93\times 10^{-4}was the K_b of ethylamine value calculated by the student.

3 0
3 years ago
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