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3 years ago

What makes the results of a scientific experiment accurate?

Chemistry

2 answers:

Dmitriy789 [7]3 years ago

6 0

Answer:

Explanation:

That's the best answer out of all of them.

Karolina [17]3 years ago

5 0

Answer: A. Having multiple trials.

Explanation:

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Answer:

complex

Explanation:

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3 years ago

What is the formula of the ion formed when tin achieves a stable electron configuration?

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3 years ago

A gas mixture contains 10.0 mole% H2O (v) and 90.0 mole % N2. The gas temperature and absolute pressure at the start of each of

Rainbow [258]

Answer: (a). T = 38.2 °C (b). V = 1.3392 cm³ (c). ii and iii

Explanation:

this is quite easy to solve, i will give a step by step analysis to solving this problem.

(a). from the question we have that;

the Mole fraction of Nitrogen, yи₂ = 0.1

Also the Mole fraction of Water, yн₂o = 0.1

We know that the vapor pressure is equal to the partial pressure because the vapor tends to condense at due point.

ρн₂o = ṗн₂o

= yн₂oP = 0.1 × 500 mmHg = 50 mmHg

from using Antoine equation, we apply the equation

logρн₂o = A - B/C+T

T = B/(A - logρн₂o) - C

= 1730.63 / (8.07131 - log 50mmHg) - 223.426

T = 38.2 °C

We have that the temperature for the first drop of liquid form is 38.2 °C

(b). We have to calculate the total moles of gas mixture in a 30 litre flask;

n = PV/RT

n = [500(mmHG) × 30L] / [62.36(mmHGL/mol K) × 323.15K] = 0.744 mol

Moles of H₂O(v) is 0.1(0.744) = 0.0744 mol

Moles of N₂ is 0.9(0.744) = 0.6696 mol

we have that the moles of water condensed is 0.0744 mol i.e the water vapor in the flask is condensed

Vн₂o = 0.0744 × 18 / 1 (g/cm³)

Vн₂o = 1.3392 cm³

Therefore, the volume of the liquid water is 1.3392 cm³

(3). (ii) and (iii)

The absolute pressure of the gas and The partial pressure of water in the gas would change if the barometric pressure drops.

cheers i hope this helps!!!!

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3 years ago

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3 years ago

Read 2 more answers

Which relationship can be used to aid in the determination of the heat absorbed by bomb calorimeter?

NARA [144]

Answer:

ΔH = $q_{p}$

Explanation:

In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.

The heat transfer is represented by

$q_{com}$ = $q_{p}$

where

$q_{p}$ = the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.

$q_{com}$ = the heat of combustion

Also, we know that the total heat change of the any system is

ΔH = ΔQ + ΔW

where

ΔH = the total heat absorbed by the system

ΔQ = the internal heat absorbed by the system which in this case is $q_{p}$

ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0

substituting into the heat change equation

ΔH = $q_{p}$ + 0

==> ΔH = $q_{p}$

5 0

3 years ago