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Valentin [98]
3 years ago
14

A sample of an ideal gas at 1.00 atm and a volume of 1.73 L was placed in a weighted balloon and dropped into the ocean. As the

sample descended, the water pressure compressed the balloon and reduced its volume. When the pressure had increased to 70.0 atm, what was the volume of the sample
Chemistry
1 answer:
kolezko [41]3 years ago
7 0

Answer:

Volume is 0.0247L

Explanation:

This question involves the use of Boyle's law which states that the volume of a given mass of gas is inversely proportional to it's pressure provided that temperature remains constant.

Mathematically,

v = k/p

K = VP

P1V1 = P2V2 = P3V3 = .........=PnVn

V = volume

p = pressure

Data;

P1 =1.0atm

V1 = 1.73L

P2 = 70atm

V2 = ?

P1V1 = P2V2

V2 = P1V1 / P2

V2 = (1.0×1.73)/70

V2 = 0.0247L

The volume of the sample is 0.0247L

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A biologist explored a remote island and discovered a new living organism. After several observations were made of the
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D) The organism was able to transport water through specialized veins
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N₂ + 3H2 → 2NH3
Rainbow [258]

Answer:

187 moles NH₃

Explanation:

To find the amount of ammonia formed, you need to multiply the given value by the mole-to-mole ratio consisting of both relevant molecules. The mole-to-mole ratio is made up of the molecules' coefficients in the balanced equation. The desired unit should be placed in the numerator of the ratio. The final answer should have 3 sig figs to reflect the lowest amount of sig figs among the given values.

1 N₂ + 3 H₂ ----> 2 NH₃
^                         ^

93.5 moles N₂          2 moles NH₃
----------------------  x  -------------------------  =  187 moles NH₃
                                    1 mole N₂

3 0
1 year ago
Read 2 more answers
Consider the following reaction: Consider the reaction 2NO(g)+Br2(g)⇌2NOBr(g) ,Kp=28.4 at 298 K In a reaction mixture at equilib
Vitek1552 [10]

Answer:  partial pressure of NOBr is 7792 atm

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

2NO(g)+Br_2(g)\rightleftharpoons 2NOBr(g)

Equilibrium constant is given as:

K_{p}=\frac{[p_{NOBr}]^2}{[p_{NO}]^2\times [p_{Br_2}]^1}

28.4=\frac{[p_{NOBr}]^2}{[(119)^2\times (151)^1}

[p_{NOBr}]=7792 atm

Partial pressure of NOBr is 7792 atm

4 0
2 years ago
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
1 year ago
Pls pls answer questions for chem final within 45 mins i will give brainliest
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#7 is D because you move the decimal 4 places
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2 years ago
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