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2 years ago

Given the balanced equation representing a reaction:

Chemistry

2 answers:

romanna [79]2 years ago

6 0

The answer should be neutralization

Murljashka [212]2 years ago

6 0

The chemical equation between an acid and a base to form salt and water is called (2) neutralization.
Hope this helps~

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1. Aluminium reacts with chlorine to produce aluminium chloride. The balanced reaction is:

Anton [14]

Answer:

Explanation:

6 0

2 years ago

when Mn2 ions are separated from the mixture, they go through a series of oxidizing and reducing steps. Write the reaction equat

goldfiish [28.3K]

Answer: hello some part of your question is missing below is the missing part

when H₂O and H₂O₂ is added to Mn(OH)₂(s) and put in water bath to dissolve

answer : attached below

Explanation:

When Mn²⁺ ions are separated from the mixture, attached below are the requires reaction equations that shows the process of separation.

Mn²⁺ ions are separated to the right of the reaction equations

8 0

2 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago

For many purposes we can treat ammonia NH3 as an ideal gas at temperatures above its boiling point of −33.°C. Suppose the temper

Keith_Richards [23]

Answer:

The new pressure will be 0.225 kPa.

Explanation:

Applying combined gas law:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1\text{ and }V_1$ are initial pressure and volume at initial temperature $T_1$ .

$P_2\text{ and }V_2$ are final pressure and volume at initial temperature $T_2$ .

We are given:

$P_1=0.29 kPa\\V_1=V\\P_2=?\\V_2=V+50\% of V=1.5 V$

$T_1=-25^oC=248.15 K$

$T _2 = 16^oC=289.15 K$

Putting values in above equation, we get:

$\frac{0.29 kPa\times V}{248.15 K}=\frac{P_2\times 1.5V}{289.15 K}$

$P_1=0.225 kPa$

Hence, the new pressure will be 0.225 kPa.

4 0

3 years ago

How does nuclear energy work?

Katena32 [7]

With computer you have a wire you connect the nuclear and the computer together that is how it works in the in the computer there is a stopwatch of 30 minutes

3 0

3 years ago