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nevsk [136]
2 years ago
11

Why does the moon's gravitational pull have more effect on Earth's tides than the sun's gravitational pull?

Chemistry
2 answers:
kvasek [131]2 years ago
8 0
It’s A. The moon is closer to the earth than the sun
BaLLatris [955]2 years ago
6 0
I believe it’s c because if you really read and close read it will make more sense
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The depletion of ozone (O3) in the stratosphere has been a matter of great concern among scientists in recent years. It is belie
Deffense [45]

<u>Answer:</u> Nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

<u>Explanation:</u>

To calculate the number of moles, we use the equation

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}   ....(1)

  • <u>For ozone:</u>

Given mass of ozone = 0.827 g

Molar mass of ozone = 48 g/mol

Putting values in above equation, we get:

\text{Moles of ozone}=\frac{0.827g}{48g/mol}=0.0172mol

  • <u>For nitric oxide:</u>

Given mass of nitric oxide = 0.635 g

Molar mass of nitric oxide = 30.01 g/mol

Putting values in above equation, we get:

\text{Moles of nitric oxide}=\frac{0.635g}{30.01g/mol}=0.0211mol

For the given chemical equation:

O_3+NO\rightarrow O_2+NO_2

By Stoichiometry of the reaction:

1 mole of ozone reacts with 1 mole of nitric oxide.

So, 0.0172 moles of ozone will react with = \frac{1}{1}\times 0.0172=0.0172moles of nitric oxide

As, given amount of nitric oxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ozone is considered as a limiting reagent because it limits the formation of product.

  • Amount of excess reagent (nitric oxide) left = 0.0211 - 0.0172 = 0.0039 moles

By Stoichiometry of the reaction:

1 mole of ozone produces 1 mole of nitrogen dioxide.

So, 0.0172 moles of ozone will react with = \frac{1}{1}\times 0.0172=0.0172moles of nitrogen dioxide

Now, calculating the mass of nitrogen dioxide from equation 1, we get:

Molar mass of nitrogen dioxide = 46 g/mol

Moles of nitrogen dioxide = 0.0172 moles

Putting values in equation 1, we get:

0.0172mol=\frac{\text{Mass of nitrogen dioxide}}{46g/mol}\\\\\text{Mass of nitrogen dioxide}=0.7912g

Hence, nitric oxide is the limiting reagent. The number of moles of excess reagent left is 0.0039 moles. The amount of nitrogen dioxide produced will be 0.7912 g.

8 0
3 years ago
A student reports the following measurements for a set of pennies: 102.0mL, 91.0 g. What is the density of the pennies.
zysi [14]

Answer:

V = 102.0 ml

m = 91.0 ml

density = ...?

density = m / V

= 91.0 / 102.0

= 0.89215 g/ml

7 0
2 years ago
PLEASE HURRY AND SHOW WORK
IgorC [24]

Answer:

The answer to your question is   P = 1.64 atm

Explanation:

Data

Volume = 2.5 x 10⁷ L

Temperature = 22°C

Pressure = ?

Moles = 1.7 x 10⁶

R = 0.082 atm L/ mol°K

Process

1.- Convert temperature to °K

Temperature = 22 + 273

                      = 295°K

2.- Use the Ideal gas law to solve this problem

                PV = nRT

- Solve for P

                P = nRT / V

- Substitution

                P = (1.7 x 10⁶)(0.082)(295) / 2.5 x 10⁷

- Simplification

                P = 41123000 / 2.5 x 10⁷

- Result

                P = 1.64 atm

3 0
3 years ago
How many grams of O₂ are required to react completely with 14.6 g of Na to form sodium oxide, Na₂O?
Bad White [126]

The balanced chemical reaction is :

O_2 + 4Na \ -> \ 2Na_2O

Number of moles of Na, n = \dfrac{14.6}{23} = 0.635 \  mol .

Now, from balance chemical reaction we can see that 1 mole of oxygen reacts with 4 moles of sodium.

So, number of moles of oxygen are :

n = \dfrac{0.635}{4}\  mole

So, amount of oxygen required is :

m = \dfrac{0.635 \times 32}{4}\  gm\\\\m = 5.08 \ gm

Therefore, 5.08 gram of oxygen will react with 14.6 gram of sodium.

7 0
2 years ago
Determine the mass of a ball with a wavelength of 3.45 × 10-34 m and a velocity of 6.55 m/s
Yuki888 [10]
<span>293 grams The formula for the wavelength of a massive particle is λ = h/p where λ = wavelength h = Plank constant (6.626070040Ă—10^â’34 J*s) p = momentum (mass times velocity) So let's solve for momentum and from there get the mass λ = h/p λp = h p = h/λ Substitute known values and solve p = 6.626070040Ă—10^â’34 J*s/3.45Ă—10^-34 m p = 1.92 J*s/m Since momentum is the product of mass and velocity, we have p = M * V p/V = M So substitute again, and solve. p/V = M 1.92 J*s/m / 6.55 m/s = M 1.92 kg*m/s / 6.55 m/s = M 1.92 kg*m/s / 6.55 m/s = M 0.293 kg = M So the mass is 293 grams</span>
6 0
3 years ago
Read 2 more answers
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