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tangare [24]
3 years ago
15

Jacob takes 7 classes, each 50.00 minutes long every day for the 5 day school week. How many hours per week does he spend in cla

ss? your answer should have for significant figures round your answer to two decimal places
Chemistry
1 answer:
Zina [86]3 years ago
5 0

Answer: He spends 29.17 hours per week

Explanation:

1 class = 50mins

7 classes = 7 x 50mins = 350mins

For 5 days in a week, we have

5 x 350mins = 1750mins

Converting to hours, we have

1750 /60 = 29.17 hours

You might be interested in
A 10.4 g sample of CaSO4 is found to contain 3.06 g of Ca and 4.89 g of O. Find the mass of sulfur in a sample of CaSO4 with a m
andrey2020 [161]

The mass of sulfur in a sample of CaSO4 with a mass of 65.8 g is 15.50g.

<h3>How to calculate mass of an element in a compound?</h3>

According to this question, a 10.4 g sample of CaSO4 is found to contain 3.06 g of Ca and 4.89 g of O.

This means that the mass of sulfur in the 10.4g of CaSO4 is 10.4g - (3.06g + 4.89g) = 10.4g - 7.95g = 2.45g

Next, we calculate the percent ratio of each element in the compound; CaSO4.

  • Ca = 3.06g/10.4g × 100 = 29.42%
  • S = 2.45g/10.4g × 100 = 23.56%
  • O = 4.89g/10.4g × 100 = 47.02%

According to this question, a sample of CaSO4 with a mass of 65.8 g is given. The mass of each element in this compound is as follows:

  • Ca = 29.42/100 × 65.8g = 19.36g
  • S = 23.56/100 × 65.8g = 15.50g
  • O = 47.02/100 × 65.8g = 30.94g

Therefore, the mass of sulfur in a sample of CaSO4 with a mass of 65.8 g is 15.50g.

Learn more about mass at: brainly.com/question/13672279

#SPJ1

5 0
2 years ago
Which of the following is equal to 2.0 liters? 200 mL 2,000 cm3 20 m3 20,000 mm3
ololo11 [35]
It equals 2000  but in this case it would be 20,000mm3
4 0
3 years ago
Solve for b, when c= a/b and a = 5.8 and c= 7.4
vova2212 [387]

Answer:

b= 42.92

Explanation:

5.8 x 7.4 = 42.92

4 0
3 years ago
Nitrogen dioxide is a red-brown gas responsible for the brown color of smog. A container of nitrogen dioxide that is at low pres
____ [38]

Answer:

Initially 1.51\times 10^{-2} moles of nitrogen dioxide were in the container .

Explanation:

Volume of the container at low pressure and at room temperature =V_1=3.4 L

Number of moles in the container = n_1

After more addition of nitrogen gas at the same pressure and temperature.

Volume of the container after addition = V_2=5.11 L

Number of moles in the container after addition=n_2=2.28\times 10^{-2} mol

Applying Avogadro's law:

\frac{Volume}{Moles}=constant (at constant pressure and temperature)

\frac{V_1}{n_1}=\frac{V_2}{n_2}

n_1=\frac{V_1\times n_2}{V_2}=\frac{3.4 L\times 2.28\times 10^{-2} mol}{5.11 L}

n_1=1.51\times 10^{-2} mol

Initially 1.51\times 10^{-2} moles of nitrogen dioxide were in the container .

8 0
3 years ago
A sample of propane(c3h8)has 3.84x10^24 H atoms.
deff fn [24]

Answer:

A) 14. 25 × 10²³ Carbon atoms

B) 34.72 grams

Explanation:

1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.

The sample has 3.84 × 10²⁴ H atoms.

If 8 atoms of Hydrogrn are present in 1 molecule of propane.

3.84 × 10²⁴ H atoms are present in

\mathfrak{ \frac{3.8 }{8} \times 10 ^{24}}

<u>= 4.75 × 10²³ molecules of Propane</u>.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

No. of Carbon atoms in 1 molecule of propane = 3

=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³

<u>= 14.25 × 10²³ </u>

<u>________________________________________</u>

<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>

= 3 × 12 + 8 × 1

= 36 + 8

= 44 g

1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.

=> 6.02 × 10²³ molecules of Propane weigh = 44 g

=> 4. 75 × 10²³ molecules of Propane weigh =

\mathsf{ \frac{44 }{6.02 \times  {10}^{23} } \times 4.75 \times  {10}^{23}  }

\mathsf{  = \frac{44 }{6.02 \times   \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}}  }

\mathsf{  = \frac{44 }{6.02 } \times 4.75   }

<u>= 34.72 g</u>

8 0
2 years ago
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