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Irina18 [472]
3 years ago
14

Question 14 of 25

Chemistry
1 answer:
Allisa [31]3 years ago
6 0

Answer:

it can be recycled over and over again

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An atom of sodium can transfer its one valence electron to an atom of...
seropon [69]

Answer:

I believe it is B,fluorine to complete the octet rule

Explanation:

4 0
3 years ago
Read 2 more answers
Im sooooooooooooooooooooooooo cufused on science
Elena-2011 [213]

Answer:

What do you need please to understand?

6 0
3 years ago
An object has a density of 10.00 g/mL. If the object has a volume of<br> 25.00 ml, what is the mass?
SVEN [57.7K]

Answer:

The answer is

<h2>250 g</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of object = 25 mL

Density = 10 g/mL

The mass of the object is

mass = 25 × 10

We have the final answer as

<h3>250 g</h3>

Hope this helps you

6 0
3 years ago
Consider the equilibrium reaction. 2 A + B − ⇀ ↽ − 4 C After multiplying the reaction by a factor of 2, what is the new equilibr
vredina [299]

Answer : The correct expression for equilibrium constant will be:

K_c=\frac{[C]^8}{[A]^4[B]^2}

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

4A+2B\rightleftharpoons 8C

The expression of K_c will be,

K_c=\frac{[C]^8}{[A]^4[B]^2}

Therefore, the correct expression for equilibrium constant will be, K_c=\frac{[C]^8}{[A]^4[B]^2}

4 0
3 years ago
Please show some work For the reaction: NO(g) + 1/2 O2(g) → NO2(g) ΔH°rxn is -114.14 kJ/mol. Calculate ΔH°f of gaseous nitrogen
uranmaximum [27]

Answer:

148.04 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

NO(g) + 1/2 O₂(g) → NO₂(g)      ΔH°rxn = -114.14 kJ/mol

We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.

ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))

ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol

ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol

ΔH°f(NO(g)) = 148.04 kJ/mol

8 0
3 years ago
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