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almond37 [142]
3 years ago
13

Given 700 ml of oxygen at 7 ºC and 106.6 kPa pressure, what volume does it take at 27ºC and 66.6 kPa pressure

Chemistry
1 answer:
o-na [289]3 years ago
6 0

Answer:

The volume is 1.2L

Explanation:

Initial volume (V1) = 700mL = 0.7L

Initial temperature (T1) = 7°C = (7 + 273.15)K = 280.15K

Initial pressure = 106.6kPa = 106600Pa

Final temperature (T2) = 27°C = (27 + 273.15)K = 300.15K

Final pressure (P2) = 66.6kPa = 66600Pa

Final volume (V2) = ?

To solve this question, we need to use combined gas equation which is a combination of Boyle's law, Charles Law and pressure law.

(P1 × V1) / T1 = (P2 × V2) / T2

solve for V2 by making it the subject of formula,

P1 × V1 × T2 = P2 × V2 × T1

V2 = (P1 × V1 × T2) / (P2 × T1)

V2 = (106600 × 0.7 × 300.15) / (66600 × 280.15)

V2 = 22397193 / 18657990

V2 = 1.2L

The final volume of the gas is 1.2L

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nikitadnepr [17]

Answer: The statement which could possibly not be true is C -" Liquid X can exist as a stable phase at 25°C, 1atm."

Explanation:

Triple point is the point where a substance co-exist as solid liquid and gas. At any point other than the triple point, the substance exist as a single phase substance.

As shown in the diagram, Liquid cannot exist as a stable phase at 1atm( below the the triple point pressure of 2atm) as the liquid can only exist beyond the pressure of triple point.

3 0
3 years ago
If an equal number of moles of reactants are used, do the following equilibrium mixtures contain primarily reactants or products
puteri [66]

<u>Answer:</u>

<u>For a:</u> The equilibrium mixture contains primarily reactants.

<u>For b:</u> The equilibrium mixture contains primarily products.

<u>Explanation:</u>

There are 3 conditions:

  • When K_{eq}>1; the reaction is product favored.
  • When K_{eq}; the reaction is reactant favored.
  • When K_{e}=1; the reaction is in equilibrium.

For the given chemical reactions:

  • <u>For a:</u>

The chemical equation follows:

HCN(aq.)+H_2O(l)\rightleftharpoons CN^-(aq.)+H_3O^+(aq.);K_{eq}=6.2\times 10^{-10}

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}

As, K_{eq}, the reaction will be favored on the reactant side.

Hence, the equilibrium mixture contains primarily reactants.

  • <u>For b:</u>

The chemical equation follows:

H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g);K_{eq}=2.51\times 10^{4}

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}

As, K_{eq}>1, the reaction will be favored on the product side.

Hence, the equilibrium mixture contains primarily products.

4 0
3 years ago
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erastovalidia [21]

Answer:

1Cu --> 1Cu+2 + 2e

2e +1S-->1S-2

Cu + S --> CuS

Explanation:

6 0
2 years ago
Which of these statements best explains why chemistry is reliable?
Dominik [7]

Answer:

It gives the same result when an experiment is repeated.

Explanation:

Below are the possible answers to the question:

<em>It is biased. </em>

<em>It cannot be verified. </em>

<em>It cannot add new evidence to existing evidence. </em>

<em>It gives the same result when an experiment is repeated.</em>

<em />

<em>The correct answer would be that </em><em>it gives the same result when an experiment is repeated.</em>

If a reaction is conducted in chemistry and certain results are obtained, once a detailed procedure of the experiment is known along with all the chemicals involved, such reaction/experiment can be repeated anywhere in the world and the same result would be obtained.

<u>The repeatability of experiments always makes the experiments to be reliable.</u> Hence, chemistry is reliable because it gives the same result without any variation when experiments are repeated under similar conditions.

7 0
3 years ago
An object was measured by a worker as
Gre4nikov [31]

Answer:

\boxed{2.8 \, \%}

Explanation:

\text{Percent error} = \dfrac{\lvert \text{Measured - Actual}\lvert}{ \text{Actual}} \times100 \, \%

Data:

Predicted = 17.4 cm

     Actual = 17.9 cm

Calculation:

\text{Percent error} = \dfrac{\lvert 17.4 - 17.9\lvert}{17.9} \times 100 \, \% \\\\= \dfrac{\lvert-0.5\lvert}{17.9} \times 100 \, \% = \dfrac{0.5}{17.9} \times 100 \, \% \\\\= 0.028 \times 100 \, \% = \textbf{2.8 \%}\\\\\text{The percent error in the measurement is } \boxed{\textbf{2.8 \%}}

6 0
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