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OLEGan [10]
3 years ago
5

Gyhgguuuyiiuyhjuhjiokjhgyhjiuh

Chemistry
2 answers:
Tasya [4]3 years ago
3 0

Answer:

gyhgguuuyiiuyhjuhjiokjhgyhjiuh

Explanation:

lol

Goryan [66]3 years ago
3 0

Explanation:

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What is the equation to find the force of a hydraulic system
Alekssandra [29.7K]

Answer:

You typically measure hydraulic pressure in pounds per square inch (psi), which is force per unit area. To calculate the force produced, multiply the pressure by the area of the hydraulic cylinder's piston in square inches. This will give you the force in pounds, which you can easily convert into tons.

Explanation:

8 0
3 years ago
The pH value of 0.1 mol dm-3 of acid U is 1. Which statement is true about acid U?
lesya [120]

Answer:

the degree of ionization in water is high

Explanation:

The term pH is defined as the negative logarithm of hydrogen ion concentration.

Hence;

pH = -log[H^+]

The pH scale shows the degree of acidity or alkalinity of a solution. A solution of pH 1 is a strong acid. A strong acid dissociates completely in solution.

Hence, acid U has a high degree of ionization in water.

3 0
3 years ago
Need help with this science question............
Brums [2.3K]

Answer:

18- H

19- D

20-G

21-A

(i think these are the answers)

Explanation:

4 0
3 years ago
Calculate the amount of heat required to raise the temperature of a 65-g sample of water from 32 ∘C to 65 ∘C. (The specific heat
dsp73

The amount of heat required is 9.0 kJ.

<em>q = mC</em>Δ<em>T </em>

Δ<em>T</em> = <em>T</em>_f – <em>T</em>_i = 65 °C – 32 °C = 33 °C

<em>q</em> = 65 g × 4.184 J·°C⁻¹g⁻¹ × 33 °C = 9000 J = 9.0 kJ

8 0
3 years ago
Read 2 more answers
Be sure to answer all parts. what are the concentrations of hso4−, so42−, and h in a 0. 31 m khso4 solution? (hint: h2so4 is a s
disa [49]

At equilibrium the concentrations of:

[HSO₄⁻] = 0.10 M;

[SO₄²⁻] = 0.037 M;

[H⁺] = 0.037 M;

There is initially very little H+ and no SO₄²⁻ in the solution. A salt is KHSO₄⁻. All KHSO₄⁻ will split apart into K⁺ and HSO₄⁻ ions. [HSO₄⁻] will initially be present at a concentration of 0.14 M.

HSO₄⁻ will not gain H⁺ to produce H₂SO₄ since H₂SO₄ is a strong acid.  HSO₄⁻ may act as an acid and lose H⁺ to form SO₄²⁻. Let the final H⁺ concentration be x M. Construct a RICE table for the dissociation of HSO₄²⁻.

R   HSO_4^-    ⇄  H^+ + SO_4^2^-

I    0.14

C   - x               +x       +x

E   0.14-x        x         x

K_a = 1.3 × 10^-^2 for HSO^-_4 . As a result,

\frac{[H^+]. [SO_4^2^-]}{HSO_4^-} = K_a

K_a is large. It is no longer valid to approximate that [HSO^-_4] at equilibrium is the same as its initial value.

\frac{x^2}{0.14-y} = 1.3 * 10^-^2

x^2+1.3*10^-^2x - 0.14 × 1.3 × 10^-^2= 0

Solving the quadratic equation for x , x \geq 0 since x represents a concentration;

                             x=0.0366538

Then, round the results to 2 significant figure;

  • [SO_4^2^-] = x = 0.037 mol. L ^-^1
  • [H^+] = x = 0.037 mol. L ^-^1
  • [HSO_4^-] = 0.14 - x = 0.10 mol. L ^-^1

Learn more about concentration here:

brainly.com/question/14469428

#SPJ4

3 0
2 years ago
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