All categories

2 years ago

What products would you expect from a reaction between hcl and koh?

1 answer:

8 0

HCl is a strong acid and KOH is a strong base. HCl dissociates to form H⁺ ions and Cl⁻ ions and KOH dissociates to give out K⁺ and OH⁻ ions.
When H⁺ and OH⁻ ions react they form H₂O molecules which are neutral.K⁺ ions and Cl⁻ ions form KCl salts.
the balanced equation for the reaction is as follows
HCl + KOH ----> KCl + H₂O
products are therefore KCl and H₂O

You might be interested in

A vessel of volume 22.4 dm3 contains 20 mol h2 and 1 mol n2 ad 273.15 k initially. All of the nitrogen reacted with sufficient h

NikAS [45]

Nitrogen combine with hydrogen to produce ammonia $\text{NH}_3$ at a $1:3:2$ ratio:

$\text{N}_2 \; (g) + 3 \; \text{H}_2 \; (g) \leftrightharpoons 2\; \text{NH}_3 \; (g)$

Assuming that the reaction has indeed proceeded to completion- with all nitrogen used up as the question has indicated. $3 \; \text{mol}$ of hydrogen gas would have been consumed while $2 \; \text{mol}$ of ammonia would have been produced. The final mixture would therefore contain

$17 \; \text{mol}$ of $\text{H}_2 \; (g)$ and
$2 \; \text{mol}$ of $\text{NH}_3 \; (g)$

Apply the ideal gas law to find the total pressure inside the container and the respective partial pressure of hydrogen and ammonia:

$\begin{array}{lll} P(\text{container}) &= & n \cdot R \cdot T / V \\ & = & (17 + 2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 1.926 \times 10^{3} \; \text{kPa} \end{array}$
$\begin{array}{lll} P(\text{H}_2) &= & n \cdot R \cdot T / V \\ & = & (17) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 1.723 \times 10^{3} \; \text{kPa} \end{array}$
$\begin{array}{lll} P(\text{NH}_3) &= & n \cdot R \cdot T / V \\ & = & (2) \; \text{mol} \times 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \\ & &\times 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=& 2.037 \times 10^{2} \; \text{kPa} \end{array}$

6 0

3 years ago

A chemist heats the block of copper as shown in the interactive, then places the metal sample in a cup of oil at 25.00 °C instea

Mazyrski [523]

When the oil is added to the heated copper, the energy in the system is

conserved.

The mass of the oil in the cup, is approximately 64.73 grams.

Reasons:

The question parameters are;

Temperature of the oil in the cup = 25.00°C

Final temperature of the oil and copper, T₂ = 27.33 °C

Specific heat of copper, c₂ = 0.387 J/(g·°C)

Specific heat capacity of oil, c₁ = 1.74 J/(g·°C)

Required:

The mass of oil in the cup.

Solution:

The mass of the copper, m₂ = 17.920 g

Temperature of copper after heating, T₂ = 65.17°C

Temperature of the copper after being placed in the cup of oil, T₂ = 27.33°C

Heat lost by copper = Heat gained by the oil

m₂·c₂·(T₂ - T₃) = m₁·c₁·(T₃ - T₁)

Therefore, we get;

17.920 × 0.387 × (65.17 - 27.33) = m₁ × 1.74 × (27.33 - 25)

262.4219136 = 4.0542·m₁

m₁ ≈ 64.73

The mass of the oil in the cup, m₁ ≈ 64.73 g

Learn more here:

brainly.com/question/21406849

Possible part of the question obtained from a similar question online, are;

The mass of the copper, m₂ = 17.920 g

Temperature of copper after heating = 65.17°C

6 0

2 years ago

2Na(s) + Cl2(g) - 2NaCl(s) + 822 kj

pychu [463]

Answer:

Is there any other part to this question? If not I'm pretty sure the answer is 205.5 kJ

Explanation:

8 0

2 years ago

Read 2 more answers

I need you to solve for me plzzzzzz

Sergeu [11.5K]

Answer:

.0556 L

Explanation:

First, convert the 1.35 M to 1.35 mol/L in order for the units to correctly cancel out.

Then, multiply (0.0725 moles Na2CO3/1) times (L/ 1.35 mol).

Finally, the answer will be .0556 L.

<h3 />

3 0

2 years ago

2. How many grams of water can be heated from 20.0 oC to 75oC using 12500.0 Joules?

kenny6666 [7]

Answer;

= 0.054 kg or 54 g

Explanation;

Using the equation; Q = mcΔT where Q is the quantity of heat transferred, m is the mass, c is specific heat of the substance, ΔT is delta T, the change in temperature.

ΔT = 75 - 20 = 55 C.

Solve the equation for m

m = Q/ cΔT

Mass = 12500 / (55 × 4200)

= 0.054 kg or 54 g

4 0

3 years ago

Read 2 more answers