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HACTEHA [7]
3 years ago
9

How do water go to the sky?

Chemistry
2 answers:
Tanya [424]3 years ago
7 0
Water goes into the sky by condensation I think
Lostsunrise [7]3 years ago
3 0
Heat from the Sun causes water<span> to evaporate from the surface of lakes and oceans.</span>
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Which of the following is an example of anaerobic respiration?
Andrews [41]
Respiring while swimming underwater
Because anaerobic respiration means respiration with no oxygen, and there is no oxygen underwater
8 0
3 years ago
5. What is the mass of 9.80 x 1023 formula units of zinc chlorate, Zn(CO3)2?
Zielflug [23.3K]

Answer:

\boxed{\text{378 g}}

Explanation:

We must convert formula units of Zn(ClO₃)₂ to moles and then to grams of Zn(ClO₃)₂.

Step 1. Convert formula units to moles

\text{Moles of Zn(ClO$_{3}$)$_{2}$}\\\\= 9.80 \times10^{23}\text{ formula units Zn(ClO$_{3}$)$_{2}$} \times \dfrac{\text{1 mol Zn(ClO$_{3}$)$_{2}$}}{6.022 \times\ 10^{23} \text{ formula units Zn(ClO$_{3}$)$_{2}$}}\\\\= \text{1.627 mol Zn(ClO$_{3}$)$_{2}$}

Step 2. Convert moles to grams

\text{Mass of Zn(ClO$_{3}$)$_{2}$}\\\\= \text{1.627 mol Zn(ClO$_{3}$)$_{2}$} \times \dfrac{\text{232.29 g Zn(ClO$_{3}$)$_{2}$}}{\text{1 mol Zn(ClO$_{3}$)$_{2}$}}\\\\= \text{378 g Zn(ClO$_{3}$)$_{2}$}\\\\\text{The mass of Zn(ClO$_{3}$)$_{2}$ is } \boxed{\textbf{378 g}}

5 0
3 years ago
A chemist has to prepare 250.0 mL of a 0.300 M Na2SO4(aq) solution. What mass, in grams, of sodium sulfate (formula mass 142.05
Sedbober [7]

The mass of sodium sulphate, Na₂SO₄, required to prepare the solution is 10.65 g

<h3>How to determine the mole of sodium sulphate Na₂SO₄</h3>
  • Volume = 250 mL = 250 / 1000 = 0.25 L
  • Molarity = 0.3 M
  • Mole of Na₂SO₄ =?

Mole = Molarity x Volume

Mole of Na₂SO₄ = 0.3 × 0.25

Mole of Na₂SO₄ = 0.075 mole

<h3>How to determine the mass of sodium sulphate Na₂SO₄</h3>
  • Molar mass of Na₂SO₄ = 142.05 g/mol
  • Mole of Na₂SO₄ = 0.075 mole
  • Mass of Na₂SO₄ =?

Mass = mole × molar mass

Mass of Na₂SO₄ = 0.075 × 142.05

Mass of Na₂SO₄ = 10.65 g

Thus, 10.65 g of Na₂SO₄ is needed to prepare the solution.

Learn more about molarity:

brainly.com/question/15370276

6 0
2 years ago
An atom of argon has a radius rar = 88 pm and an average speed in the gas phase at 25°C of 172 m/s.
Rudik [331]

Answer:

1.2* 10³ rNe.

Explanation:

Given speed of neon=350 m/s

Un-certainity in speed= (0.01/100) *350 =0.035 m/s

As per heisenberg uncertainity principle

Δx*mΔv ≥\frac{h}{4\pi }

4π

h

..................(1)

mass of neon atom =\frac{20*10^{-3} }{6.22*10^{-23} } =3.35*10^{-26} kg

6.22∗10

−23

20∗10

−3

=3.35∗10

−26

kg

substituating the values in eq. (1)

Δx =4.49*10^{-8}10

−8

m

In terms of rNe i.e 38 pm= 38*10^{-12}10

−12

Δx=\frac{4.49*10^{-8} }{38*10^{-12} }

38∗10

−12

4.49∗10

−8

=0.118*10^{4}10

4

* (rNe)

=1.18*10³ rN

= 1.2* 10³ rNe.

Explanation:

This is the answer

7 0
3 years ago
Calculate the number of pounds of CO2CO2 released into the atmosphere when a 22.0 gallon22.0 gallon tank of gasoline is burned i
Gnesinka [82]

Answer:

391.28771 pounds of carbon-dioxide was released into the atmosphere when 22.0 gallon tank of gasoline is burned in an automobile engine.

Explanation:

Density of the gasoline ,d= 0.692 g/mL

Volume of gasoline in an tanks,V = 22.0 gallons = 83,279.02 mL

Let mass of the gasoline be M

Density= \frac{Mass}{Volume}

M = V × d = 83,279.02 mL × 0.692 g/mL=57,629.081 g

Given that gasoline is primarily octane.

2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O

Mass of octane burnt in the tank = M = 57,629.081 g

Moles of octane =\frac{57,629.081 g}{114.08g/mol}=505.1637 mol

According to reaction, 2 moles of octane gives 16 moles of carbon-dioxide.

Then 505.1637 mol of octane will give:

\frac{16}{2}\times 505.1637 mol=4,041.3100 mol of carbon-dioxide

Mass of 4,041.3100 mol of carbon-dioxide:

4,041.3100 mol × 44.01 g/mol = 177,858.05 g

Mass of carbon-dioxide produced in pounds = 391.28771 pounds

391.28771 pounds of carbon-dioxide was released into the atmosphere when 22.0 gallon tank of gasoline is burned in an automobile engine.

3 0
3 years ago
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