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1 year ago

What is the average atomic mass of an element that has 3 isotopes with the data below?

Chemistry

1 answer:

Bezzdna [24]1 year ago

3 0

Answer:

How do you calculate the average atomic mass of three isotopes?

To calculate the average atomic mass, multiply the fraction by the mass number for each isotope, then add them together. Whenever we do mass calculations involving elements or compounds (combinations of elements), we always use average atomic masses

Explanation:

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the atomic number is the number of protons, but in a neutral atom it is also the number of _____. A. none of these B. electrons

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Answer:

Electrons

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2 years ago

Soil that are light in color have a great amount of humus in<br> them. True or False

GaryK [48]

Answer:

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1 year ago

A solution is made by adding 29.1 mL of concentrated perchloric acid ( 70.5 wt% , density 1.67 g/mL ) to some water in a volumet

Lera25 [3.4K]

Answer:

The concentration of the solution is 1.364 molar.

Explanation:

Volume of perchloric acid = 29.1 mL

Mass of the solution = m

Density of the solution = 1.67 g/mL

$m=1.67 g/mL\times 29.1 mL=48.597 g$

Percentage of perchloric acid in 48.597 solution :70.5 %

Mass of perchloric acid in 48.597 solution :

= $\frac{70.5}{100}\times 48.597 = 34.261 g$

Moles of perchloric acid = $\frac{34.261 g}{100.46 g/mol}=0.3410 mol$

In 29.1 mL of solution water is added and volume was changed to 250 mL.

So, volume of the final solution = 250 mL = 0.250 L (1 mL = 0.001 L)

$Molarity=\frac{Moles}{Volume (L)}$

$=\frac{0.3410 mol}{0.250 L}=1.364 M$

The concentration of the solution is 1.364 molar.

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2 years ago

Choose all the answers that apply.

Anni [7]

Answer:

I believe it's the lowest portion of the atmosphere

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2 years ago

What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter

Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole$

Now put all the given values in the above formula, we get:

$44.5kJ/mol=\frac{q}{0.038mol}$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

Now put all the given values in the above formula, we get:

$1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

4 0

2 years ago