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2 years ago

2. A lab group turned in their experiment and had measured the mass of their sample

Chemistry

1 answer:

alisha [4.7K]2 years ago

3 0

Answer:

1.61%.

Explanation:

The following data were obtained from the question:

Measured mass = 3.15 g.

Actual mass = 3.10 g

Percentage error =.?

The percentage error of the group can be obtained by using the following equation below:

Percentage error = |Actual mass – Measured mass| / Actual mass × 100

Percentage error = |3.10 – 3.15| /3.10 × 100

Percentage error = 0.05/3.1 × 100

Percentage error = 5/3.1

Percentage error = 1.61%

Therefore, the percentage error of the group is 1.61%.

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atmospheric pressure at elevations of 8000 feet averages about 0.72 atmospheres. Would a cabin pressurized at 500 mm hg meet fed

NeTakaya

Answer:

Yes, but it must be kept at that value and do not let it to decrease more.

Explanation:

Hello.

In this case, in order to substantiate whether the cabin meet the federal standards, we need to convert the 500 mmHg to atm and compare the result with 0.72 atm by knowing that 1 atm equals 760 mmHg:

$500mmHg*\frac{1atm}{760mmHg} \\\\=0.66atm$

Thus, since 0.66 atm is 0.06 atm away from the federal standard we can infer that it may meet the federal standard, however, it would not be recommended to let the pressure decrease more than that.

8 0

2 years ago

A buffer solution is composed of 1.00 mol of acid and 2.25 mol of the conjugate base. If the p K a of the acid is 4.90 , what is

Gemiola [76]

<u>Answer:</u> The pH of the buffer is 5.25

<u>Explanation:</u>

Let the volume of buffer solution be V

We know that:

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution}}$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{conjugate base}]}{[acid]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of weak acid = 4.90

$[\text{conjugate base}]=\frac{2.25}{V}$

$[acid]=\frac{1.00}{V}$

pH = ?

Putting values in above equation, we get:

$pH=4.90+\log(\frac{2.25/V}{1.00/V})\\\\pH=5.25$

Hence, the pH of the buffer is 5.25

4 0

2 years ago

Please help me with this homework

ella [17]

Answer:

Density is the correct answer choice.

5 0

3 years ago

Read 2 more answers

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$

Finally, we compute the percent yield with the obtained 2.10 g:

$Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%$

Best regards.

7 0

2 years ago

Describe the valence electrons in nitrogen and how it could bond to other atoms.

Trava [24]

Answer:

Each nitrogen molecule consists of two atoms of nitrogen that are bonded by a triple covalent bond. This is a direct consequence of the fact that each nitrogen atom has 5 valence electrons. Each atom can thus complete its octet by sharing three electrons.

Explanation:

Sorry i had to look it up i didn't know this answer

Please mark brainliest:)

6 0

2 years ago