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2 years ago

The image below shows a certain type of global wind:

Chemistry

1 answer:

forsale [732]2 years ago

6 0

Answer:

Number 3

It is the way the earth stay a good tempercher for us.

Explanation:

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HELP I WILL MARK YOU AS THE BRAINLIEST ANSWER

borishaifa [10]

Hey there! Here is the answer:

B.) decrease; increase

Here's why:

As you move from left to right across a period on the periodic table the size of an atom will decrease. As you move from top to bottom down a group on the periodic table the size of an atom will increase.

I really hope this answer helps you out! :)

8 0

3 years ago

What was the biggest challenge you faced in getting to where you are today and how did you overcome it?

Alchen [17]

It’s all based on you but yours could be anything hard you went through and how you overcame it. Also say how it made you the person you are today

4 0

3 years ago

N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?

just olya [345]

Answer:

$\large \boxed{\text{197.4 g}}$

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 28.01 17.03

N₂ + 3H₂ ⟶ 2NH₃

m/g: 240.0

(a) Moles of NH₃

$\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}$

(b) Moles of N₂

$\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}$

(c) Mass of N₂

$\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}$

7 0

3 years ago

Read 2 more answers

Helppppp with this pleasee

vazorg [7]

Answer:

A. 1, 2, 5

Explanation:

Count the number of Ns in the formula.

- Hope that helped! Please let me know if you need a further explanation.

3 0

3 years ago

Suppose a 250. mL flask is filled with 0.30 mol of N_2 and 0.70 mol of NO. The following reaction becomes possible:N_2(g) +O2 →

Inessa [10]

Answer:

0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

[N₂] = (0.30 + x)/0.250

[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = $\frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }$

7.70 = (0.70-2x)²/[(0.30+x)*x]

7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

3.7x² + 5.11x - 0.49 = 0

Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

Thus,

[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

3 0

3 years ago