All categories

2 years ago

how many grams of solid silver nitrate would you need to prepare 200.0 mL of a 0.150 M AgNO3 solution

1 answer:

4 0

0.150 M AgNO3 = x mol / 0.200 Liters
x mol = 0.03 mol AgNO3
0.03 mol AgNO3 (169.9g AgNO3 / 1 mol AgNO3) We are converting moles to grams here with stoichiometry.

Final answer = 5.097 grams, but if you want it in terms of sig figs then it is 5.09 grams.

You might be interested in

Which of the following would be expected to have the highest viscosity? CH3CH2OH, HOCH2OH, CH3CH2CH3

zaharov [31]

Answer: Out of the given options $HOCH_{2}OH$ is expected to have the highest viscosity.

Explanation:

The resistance occurred in the flow of a liquid substance is called viscosity.

More stronger is the intermolecular forces present in a substance more will be its resistance in its flow. Hence, more will be its viscosity.

For example, $HOCH_{2}OH$ has strong intermolecular hydrogen bonding than the one's present in $CH_{3}CH_{2}OH$ and $CH_{3}CH_{2}CH_{3}$ . This is because two-OH groups are present over here.

Thus, we can conclude that out of the given options $HOCH_{2}OH$ is expected to have the highest viscosity.

5 0

2 years ago

In a living organism, what is an organ? functional unit, or building block, of all organisms; smallest unit that can carry on th

ra1l [238]

Answer: D. a structure that provides a specific service within a cell

Explanation:

7 0

2 years ago

How many atoms are equal to 1.5 moles of hellium

ANTONII [103]

Answer:

There are 1.8×1024 atoms in 1.5 mol HCl

Explanation:

6 0

2 years ago

Read 2 more answers

A sample of gas contains 0.1700 mol of OF2(g) and 0.1700 mol of H2O(g) and occupies a volume of 19.5 L. The following reaction t

andreev551 [17]

Answer: The volume of the sample after the reaction takes place is 29.25 L.

Explanation:

The given reaction equation is as follows.

$OF_{2}(g) + H_{2}O(g) \rightarrow O_{2}(g) + 2HF(g)$

So, moles of product formed are calculated as follows.

$\frac{3}{2} \times 0.17 mol \\= 0.255 mol$

Hence, the given data is as follows.

$n_{1}$ = 0.17 mol, $n_{2}$ = 0.255 mol

$V_{1}$ = 19.5 L, $V_{2} = ?$

As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$

Substitute the values into above formula as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{19.5 L}{0.17 mol} = \frac{V_{2}}{0.255 mol}\\V_{2} = \frac{19.5 L \times 0.255 mol}{0.17 mol}\\= \frac{4.9725}{0.17} L\\= 29.25 L$

Thus, we can conclude that the volume of the sample after the reaction takes place is 29.25 L.

8 0

2 years ago

How do I calculate the molecular weight of Mg(OH)2?

vodomira [7]

The molecular weight of Mg(OH)2 : 58 g/mol

<h3>Further explanation</h3>

Given

Mg(OH)2 compound

Required

The molecular weight

Solution

Relative atomic mass (Ar) of element : the average atomic mass of its isotopes

Relative molecular weight (M) : The sum of the relative atomic mass of Ar

M AxBy = (x.Ar A + y. Ar B)

So for Mg(OH)2 :

= Ar Mg + 2 x Ar O + 2 x Ar H

= 24 g/mol + 2 x 16 g/mol + 2 x 1 g/mol

= 24 + 32 + 2

= 58 g/mol

5 0

2 years ago