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Serggg [28]
3 years ago
15

Convert 4.8 moles of calcium carbonate to particles.

Chemistry
1 answer:
Iteru [2.4K]3 years ago
8 0

Answer:

<h2>2.89 × 10²⁴ particles</h2>

Explanation:

The number of particles can be found by using the formula

N = n × L

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is 6.02 × 10²³ entities

We have

N = 4.8 × 6.02 × 10²³

We have the final answer as

<h3>2.89 × 10²⁴ particles</h3>

Hope this helps you

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Which of the following statements is true?
Fittoniya [83]

Answer:

Es la tercera por qué cuando el material varía va cambiando de estado

Explanation:

8 0
2 years ago
Convert 8.50 moles Ca to atoms
Wewaii [24]

8.50 moles is equal to 5.1187×10²⁴ atoms of Ca.

<u>Explanation:</u>

We have to multiply the moles of Ca by the Avogadro's number:

= 6.022×10²³

So the number of atoms:

= 8.5 moles × 6.022×10²³atoms / mol

= 5.1187×10²⁴ atoms

Hence the 8.50 moles is equal to 5.1187×10²⁴ atoms of Ca.

4 0
3 years ago
7. Explain how magnetic potential energy can be transformed into kinetic energy. (1<br> point)
Arisa [49]
The potential energy by the magnetic field can turn into kinetic energy once the field is moving from the S pole to the N pole when it reaches the N pole it is potential energy when it exits the S pole it is kinetic energy.
5 0
2 years ago
When the following reaction comes to equilibrium, will the concentrations of the reactants or products be greater? Does the answ
Nutka1998 [239]

Answer:

At equilibrium, the concentration of the reactants will be greater than the concentration of the products. This does not depend on the initial concentrations of the reactants and products.

Explanation:

The value of Kc gives us an idea of the extent of the reaction. A big Kc (Kc > 1) means that in the equilibrium there are more products than reactants, and the opposite happens for a small Kc (Kc < 1). The equilibrium is reached no matter what the initial concentrations are.

The value of the equilibrium constant is relatively SMALL; therefore, the concentration of reactants will be GREATER THAN the concentration of products. This result is INDEPENDENT OF the initial concentration of the reactants and products.

4 0
3 years ago
The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

K_a\times K_b=K_w

K_b for sodium formate is K_b=\frac {K_w}{K_a}

Given that:

K_a of formic acid = 1.8\times 10^{-4}

And, K_w=10^{-14}

So,

K_b=\frac {10^{-14}}{1.8\times 10^{-4}}

K_b=5.5556\times 10^{-11}

Concentration = 0.35 M

HCOONa    ⇒     Na⁺ +    HCOO⁻

Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

At t=0                              0.35                            -              -

At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

5.5556\times 10^{-11}=\frac {x^2}{0.35-x}

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

5 0
3 years ago
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