What is the molarity when 50 grams of NaCl are added to 0.5 L of water

A sample of neon gas occupies 105 L at 27°C under a pressure of

Viefleur [7K]

Answer: Volume occupied by given neon sample at standard condition is 123.84 L.

Explanation:

Given: $V_{1}$ = 105 L, $T_{1} = 27^{o}C = (27 + 273) K = 300 K$ , $P_{1}$ = 985 torr

At standard conditions,

$T_{2}$ = 273 K, $P_{2}$ = 760 K, $V_{2}$ = ?

Formula used to calculate the volume is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{985 torr \times 105 L}{300 K} = \frac{760 torr \times V_{2}}{273 K}\\V_{2} = \frac{94116.75}{760} L\\= 123.84 L$

Thus, we can conclude that volume occupied by given neon sample at standard condition is 123.84 L.

8 0

3 years ago

How many molecules are in a drop of water weighing 0.05 g?

Scilla [17]

Answer:

0.05 g of H2O contain molecules

7 0

3 years ago

A geochemist measures the concentration of salt dissolved in Lake Parsons and finds a concentration of 21.0 gL⁻¹. The geochemist

dedylja [7]

Answer:

The percentage of Lake Parsons which has evaporated since it became isolated is 68.24%.

Explanation:

The concentration of salt dissolved in Lake Parson = 21.0 g/L

= In 1 L of lake Parson water = 21.0 g of salt

The concentration of salt in several nearby non-isolated lakes = 6.67 g/L

The salt concentration in Lake Parsons before it became isolated = 6.67 g/L

In 1 L of lake water = 6.67 g of salt

If 1 L of lake Parson water has 6.67 g of salt. Then 21.0 grams of salt will be in ;

$\frac{1}{6.67}\times 21.0 L = 3.15 L$

Water evaporated = 3.15 L - 1 L = 2.15 L

The percentage of Lake Parsons which has evaporated since it became isolated:

$\frac{2.15 L}{3.15 L}\times 100=68.25\%$

5 0

4 years ago

A student determines the iron(III) content of a solution by first precipitating it as iron(III) hydroxide, and then decomposing

Semmy [17]

Answer:

3.3535 g

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For iron(III) nitrate :

Molarity = 0.404 M

Volume = 52.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 52.0×10⁻³ L

Thus, moles of iron(III) nitrate :

$Moles=0.404 \times {52.0\times 10^{-3}}\ moles$

Moles of iron(III) nitrate = 0.021 moles

1 mole of iron(III) nitrate forms 1 mole of iron(III) hydroxide which further forms 1 mole of iron(III) oxide.

Thus, moles of iron(III) oxide formed from 0.021 moles of iron(III) nitrate = 0.021 moles

Also, molar mass of iron(III) oxide = 159.69 g/mol

So, mass of iron(III) oxide = 0.021 moles × 159.69 g/mol = 3.3535 g

6 0

3 years ago

If the atomic number of Chlorine atom is 17 and atomic mass is 35.5u, compute the number of protons, electrons and neutrons.

lord [1]

Answer:

Protons= 17, Electrons= 17, Neutrons= 18

Explanation:

The atomic number (Z) corresponds to the number of electrons that is equal to the number of protons (17). The mass number (A) corresponds to the sum of protons and neutrons present in the nucleus (35). Therefore, we calculate the number of neutrons: 35-17 = 18

5 0

3 years ago