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ANEK [815]
3 years ago
5

I need help please. It’s really urgent . thank you

Chemistry
1 answer:
wel3 years ago
3 0
2nd one, the molecules of the food in the oven are moving faster than the molecules of the food in the freezer.


Even if the food is frozen, the molecules are still moving because they have energy, but when the food is hot, the molecules move even faster.
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A sample of vinegar was found to have an acetic acid concentration of 0.8846 m. What is the acetic acid % by mass? Assume the de
jenyasd209 [6]

Answer:

5.3%

Explanation:

Let the volume be 1 L

volume , V = 1 L

use:

number of mol,

n = Molarity * Volume

= 0.8846*1

= 0.8846 mol

Molar mass of CH3COOH,

MM = 2*MM(C) + 4*MM(H) + 2*MM(O)

= 2*12.01 + 4*1.008 + 2*16.0

= 60.052 g/mol

use:

mass of CH3COOH,

m = number of mol * molar mass

= 0.8846 mol * 60.05 g/mol

= 53.12 g

volume of solution = 1 L = 1000 mL

density of solution = 1.00 g/mL

Use:

mass of solution = density * volume

= 1.00 g/mL * 1000 mL

= 1000 g

Now use:

mass % of acetic acid = mass of acetic acid * 100 / mass of solution

= 53.12 * 100 / 1000

= 5.312 %

≅ 5.3%

3 0
3 years ago
Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the fi
Savatey [412]

The question is incomplete, the complete question is;

Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the first heating, your crucible and contents weighs 17.51 g. After the second heating, your crucible and contents weighs 17.50 g.

What is the theoretical yield of sodium carbonate?

What is the experimental yield of sodium carbonate?

What is the percent yield for sodium carbonate?

Which errors could cause your percent yield to be falsely high, or even over 100%?

Answer:

See Explanation

Explanation:

We have to note that water is driven away after the second heating hence we are concerned with the weight of the pure dry product.

Hence;

From the reaction;

2 NaHCO3 → Na2CO3(s) + H2O(l) + CO2(g)

Number of moles of  sodium bicarbonate = 18.56 - 15.98 = 2.58 g/87 g/mol

= 0.0297 moles

2 moles of sodium bicarbonate yields 1 mole of sodium carbonate

0.0297 moles of 0.015 moles  sodium bicarbonate yields 0.0297 * 1/2 = 0.015 moles

Theoretical yield of sodium carbonate = 0.015 moles * 106 g/mol = 1.59 g

Experimental yield of sodium bicarbonate = 17.50 g - 15.98 g = 1.52 g

% yield = experimental yield/Theoretical yield * 100

% yield = 1.52/1.59 * 100

% yield = 96%

The percent yield may exceed 100% if the water and CO2 are not removed from the system by heating the solid product to a constant mass.

5 0
3 years ago
Question 2(Multiple Choice Worth 4 points) (01.01 MC) Dominic and Eva are using the same type of stopwatches to measure the time
LekaFEV [45]
The results of Dominic and Eva's experiment is unreliable and can lead to a pseudoscientific claim primarily because they did not repeat their tests multiple times. Although, they used the same type of stopwatch in recording the time it takes for the chemical reaction to occur, they have different reactions times. Thus, it would have been better if they conducted several trials then obtained the average of their results.    
6 0
3 years ago
Read 2 more answers
Based on the Lewis/electron dot representation
timama [110]

Answer:

B or C think so

4 0
2 years ago
What is the mass in g of 7.009 x 10^21 molecules of Pl3?
ycow [4]
I have completed the problem

4 0
3 years ago
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