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3 years ago

An aqueous solution contains three sulfate ions for every two aluminum ions. What is the solute?

Chemistry

1 answer:

german3 years ago

6 0

The aqueous solution contains $Al ^{+3}$ and $SO4 ^{-2}$ ions. The solute is Al2(SO4)3.

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A car with a mass of 1,000 kg is moving at a speed of 20 m/s. What is the car's kinetic energy?

shtirl [24]

The answer would be 200000 J. the equation for kinetic energy is 1/2 mass times velocity squared. 1/2 of 1,000 is 500. and 20*20 is 400. So, multiply 400 by 500, and that gives you your answer,

7 0

2 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago

Which statement correctly describes two forms of oxygen, O2 and O3?

fgiga [73]

The two forms of oxygen, O2 and O3 is "<span>They have different molecular structures and different properties."</span>

7 0

3 years ago

Read 2 more answers

Convert 1000mg=__g 1L=__mL 160cm=__mm 1.4km=__m 109g=__kg 250m=__km 80cm=__m 75mL=__L 5.6m=__cm 6.5g=__mg 170.4m=__cm 564Dg=__g

Snowcat [4.5K]

Answer:

1000mg= 1g

1L= 1000 mL

160cm = 1600mm

1.4km= 1400m

109 g = 0.109kg

250m= 0.250 km

80cm= 0.8 m

75mL= 0.075L

5.6m= 560 cm

6.5g= 6500mg

170.4m= 17040 cm

564 Dg = 5640 g

58 dg = 5800 mg

600 L= 0.6 KL

0.0923Km= 92300 mm

Explanation:

1 mg = 1x10⁻³ g

1 g = 1000 mg

1 g = 10 dg

1 g = 1x10⁻³ kg

1 Dg = 10 g

1 dg = 100 mg

1 L = 1000 mL

1 L = 1x10⁻³ KL

1 mL = 1x10⁻³ L

1 km = 1000 m

1 km = 1x10⁶ mm

1 m = 1x10⁻³ km

1 cm = 1x10⁻² m

1 cm = 10 mm

5 0

3 years ago

If the temperature changes by 100k by how much does it change in °c?

telo118 [61]

Answer:

-173.15°C

Explanation:

Given data:

Temperature changes = 100 K

Temperature changes in degree Celsius = ?

Solution:

Kelvin and degree Celsius both are units of temperature.

In order to convert the degree Celsius to kelvin following equation is used.

Temperature in degree Celsius +273.15

For example;

100°C to kelvin = 100+273.15 = 373.15 K

To convert the kelvin into degree Celsius:

100K - 272.15 = -173.15°C

3 0

3 years ago