Answer: 1) A solid product of a chemical reaction that is in aqueous form.
Lead fluoride hope this helps
Answer:
The answer to your question is: 0.028 kg of NO2
Explanation:
Data
3.7 x 10²⁰ molecules of NO2 in kg
MW of NO2 = 14 + (16 x 2) = 14 + 32 = 46 kg
1 mol of NO2 --------------------- 6.023 x 10 ²³ molecules
x --------------------- 3.7 x 10²⁰ molecules
x = 3.7 x 10²⁰ x 1 / 6.023 x 10 ²³
x = 0.00061 mol
1 mol of NO2 --------------------- 46 kg of NO2
0.00061 mol ------------------ x
x = 0.00061 x 46/1
x = 0.028 kg of NO2
Answer: There are 0.006 moles of acid in the flask.
Explanation:
Given:
= 21.35 mL,
= 0.150 M
= 25.0 mL,
= ?
Formula used to calculate molarity of
is as follows.

Substitute the values into above formula as follows.

As molarity is the number of moles of a substance present in a liter of solution.
Total volume of solution = 
= 21.35 mL + 25.0 mL
= 46.36 mL (1 mL = 0.001 L)
= 0.04636 L
Therefore, moles of acid required are calculated as follows.

Thus, we can conclude that there are 0.006 moles of acid in the flask.
Answer:
Approximately 75%.
Explanation:
Look up the relative atomic mass of Ca on a modern periodic table:
There are one mole of Ca atoms in each mole of CaCO₃ formula unit.
- The mass of one mole of CaCO₃ is the same as the molar mass of this compound:
. - The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element:
.
Calculate the mass ratio of Ca in a pure sample of CaCO₃:
.
Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be
of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio
:
.
In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:
.