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2 years ago

What is the "average atomic mass" of this set of pennies.

Chemistry

1 answer:

natulia [17]2 years ago

7 0

Answer:

A. 2.74g is the correct answer

Explanation:

<h3>Greetings!</h3>

$average \: mass = \frac{(mass \: pre - 1982 \times abundance) + (mass \: post - 1982 \times abundance}{100 } \\ ave = \frac{ (3.1g \times 40.0) + (2.5g \times 60.0)}{100} \\ ave = \frac{124 + 150}{100} \\ ave = \frac{274}{10} = 2.74g$

where, the sum of abundance always have to be 100%

Thus, 40.0 +60.0= 100.0

You might be interested in

Equilibrium is established between a liquid and its vapor when A. the rate of evaporation equals the rate of condensation. B. eq

Advocard [28]

Answer:

Explanation:

the rate of evaporation equals the rate of condensation.

6 0

3 years ago

How many moles are in 1.52908x1024 molecules of Potassium phosphate (K3(PO4))?

lara31 [8.8K]

Answer:

<h2>2.54 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

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

From the question we have

$n = \frac{1.52908 \times {10}^{24} }{6.02 \times {10}^{23} } \\$

We have the final answer as

<h3>2.54 moles</h3>

Hope this helps you

4 0

2 years ago

Un tecnico di laboratorio deve preparare una soluzione di carbonato di sodio decaidrato, Na2CO3⋅ 10 H2O per eseguire alcune anal

fiasKO [112]

Answer:

1. 3.70 g Na₂CO₃·10H₂O

2. 50.0 mL of the first solution

Explanation:

1. Prepare the solution

(a) Calculate the molar mass of Na₂CO₃·10H₂O

$\begin{array}{rrr}\textbf{Atoms} &\textbf{M}_{\textbf{r}} & \textbf{Mass/u}\\\text{2Na} & 2\times22.99 & 45.98\\\text{1C} & 1\times 12.01 & 12.01\\\text{13O}&13 \times16.00 & 208.00\\\text{20H}&20 \times 1.008 & 20.16\\&\text{TOTAL =} & \mathbf{286.15}\\\end{array}$

The molar mass of Na₂CO₃·10H₂O is 286.15 g/mol.

(b) Calculate the moles of Na₂CO₃·10H₂O

$\text{Moles of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\= \text{0.250 L solution} \times \dfrac{\text{0.0500 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 L solution}}\\\\= \text{0.0125 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}$

(c) Calculate the mass of Na₂CO₃·10H₂O

$\text{Mass of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O }\\= \text{0.012 50 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O } \times \dfrac{\text{296.15 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}\\\\= \text{3.70 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\\text{You need $\large \boxed{\textbf{3.70 g}}$ of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}$

2. Dilute the solution

We can use the dilution formula to calculate the volume needed.

V₁c₁ = V₂c₂

Data:

V₁ = ?; c₁ = 0.0500 mol·L⁻¹

V₂ = 100 mL; c₂ = 0.0250 mol·L⁻¹

Calculation:

$\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\V_{1}\times \text{0.0500 mol/L} & = & \text{100 mL} \times\text{0.0250 mol/L}\\0.0500V_{1}& = & \text{2.500 mL}\\V_{1}&=& \dfrac{\text{2.500 mL}}{0.0500}\\\\& = & \text{50.0 mL}\\\end{array}\\\text{You need $\large \boxed{\textbf{50.0 mL}}$ of the concentrated solution.}$

7 0

3 years ago

In the following list, only is not an example of a chemical reaction

Sladkaya [172]

There is no pictu-res...sss

4 0

3 years ago

If two gases are present in a container, the total pressure in the container is equal to

dmitriy555 [2]

Answer: the sum of the partial pressures of the individual gases.

Explanation:

According to Dalton's Law of partial pressure, the total pressure of a mixture of gases is equal to the sum of the partial pressures which each individual gas would exert if it were confined alone in the volume occupied by the mixture.

Hence, Ptotal = P1+ P2

where Ptotal is the total pressure

P1 and P2 are the partial pressures exerted seperately by the individual gases 1 and 2 that make up the mixture.

8 0

3 years ago