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Eddi Din [679]
3 years ago
14

What happens to the concentration of a solution as more solvent is added, but the amount of solute remains the same?

Chemistry
1 answer:
elena55 [62]3 years ago
6 0

Answer: CCCCCCCCCCCCCCCCCCCCCc

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A gas occupies 40.0 mL at 150 K. What volume does it occupy at 300 K,
nlexa [21]

Answer:

.08 L or 80 ml

Explanation:

Use the equation V/t = V/t.

.04L / 150K = V / 300K

.04 / 150 * 300 = V

.08 L or 80 ml

6 0
3 years ago
What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91
Vikentia [17]

Answer:  A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L

Temperature = 91^{o}C = (91 + 273) K = 364 K

As molar mass of O_2 is 32 g/mol. Hence, the number of moles of O_2 are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

PV = nRT\\P  \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

4 0
3 years ago
If a 100. -g sample of a hydrated compound contains 37.07-g sodium, 48.39-g carbonate and 14.54-g water, find the empirical form
Mumz [18]

he required empirical formula based on the data provided is Na2CO3.H2O.

<h3>What is empirical formula?</h3>

The term empirical formula refers to the formula of a compound which shows the ratio of each specie present.

We have the following;

Mass of sodium = 37.07-g

Mass of carbonate = 48.39 g

Mass of water = 14.54-g

Number of moles of sodium = 37.07-g/23 g/mol = 2 moles

Number of moles of carbonate = 48.39 g/61 g/mol = 1 mole

Number of moles of water = 14.54/18 g/mol = 1 mole

The mole ratio is 2 : 1: 1

Hence, the required empirical formula is Na2CO3.H2O

Learn more about empirical formula : brainly.com/question/11588623

3 0
2 years ago
Does vinegar conduct? yes or no
uranmaximum [27]
Im guessing and i think the answer is yes.
8 0
2 years ago
Read 2 more answers
Pls help me on this question????
astraxan [27]

Explanation:

The first one is series circuit as it is joined one after another while the second one is parallel circuit joined in a parallel way. Ok the line just like this --------- is the wire , the sign like this | | is the battery and the circles with cross are the bulbs

Hope this helps :) Have a great day

7 0
3 years ago
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