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Sergeeva-Olga [200]
3 years ago
11

You need to make an aqueous solution of 0.192 M barium sulfide for an experiment in lab, using a 500 mL volumetric flask. How mu

ch solid barium sulfide should you add?
Chemistry
1 answer:
shutvik [7]3 years ago
7 0

Answer:

Explanation:

Molecular weight of barium sulphide = 169

500 mL of .192 M barium sulfide = .5 x .192 moles of barium sulphide

= .096 moles of barium sulfide

= .096 x 169 gram of barium sulfide

= 16.22 grams of barium sulfide .

We shall have to add 16.22 gram .

You might be interested in
How many atoms are there in 8.88 g Si?
Mariana [72]

Answer:

\boxed {\boxed {\sf 1.90 \times 10^{23} \ atoms \ Si}}

Explanation:

We are asked to find how many atoms are in 8.88 grams of silicon.

<h3>1. Grams to Moles </h3>

First, we convert grams to moles. We use the molar mass or the mass of 1 mole of a substance. These values are found on the Periodic Table as they are equal to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up silicon's molar mass.

  • Si:  28.085 g/mol

We will convert using dimensional analysis. Set up a conversion factor with the molar mass.

\frac { 28.085 \ g \  Si}{1 \ mol \ Si}

We are converting 8.88 grams of silicon to moles, so we multiply by this value.

8.88 \ g \ Si *\frac { 28.085 \ g \  Si}{1 \ mol \ Si}

Flip the fraction so the units of grams of silicon cancel.

8.88 \ g \ Si *\frac{1 \ mol \ Si} { 28.085 \ g \  Si}

8.88  *\frac{1 \ mol \ Si} { 28.085 }

\frac {8.88} { 28.085 } \ mol \ Si

0.316183015845 \ mol \ Si

<h3>2. Moles to Atoms </h3>

Next, we convert moles to atoms. We use Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of silicon.

Set up another conversion factor.

\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}

Multiply by the number of moles we calculated.

0.316183015845\ mol \ Si *\frac {6.022 \times 10^{23} \ atoms \ Si}{1 \ mol \ Si}

The units of moles of silicon cancel.

0.316183015845 * \frac {6.022 \times 10^{23} \ atoms \ Si}{1}

0.316183015845 * {{6.022 \times 10^{23} \ atoms \ Si}

1.90405412 \times 10^{23} \ atoms \ Si

<h3>3. Significant Figures</h3>

The original measurement of 8.88 grams has 3 significant figures, so our answer must have the same.

For the number we calculated, that is the hundredth place. The 4 in the thousandth place tells us to leave the 0 in the hundredth place.

1.90 \times 10^{23} \ atoms \ Si

<u>8.88 grams of silicon contains 1.90 ×10²³ atoms of silicon.</u>

6 0
3 years ago
If the earth is about 13000 light years away from the outermost edge of the milky way galaxy, what is the total diameter of the
xeze [42]

Answer: The Milky Way is a barred spiral galaxy with a diameter between 150,000 and 200,000 light-years (ly).

3 0
3 years ago
SO
Burka [1]

Answer:

\large \boxed{\text{E) 721 K; B) 86.7 g}}

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL;  V₂ = 435 mL

T₁ = 355 K;     T₂ = ?

b) Calculation

\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}

7 0
3 years ago
Nurse Antonio measured out 7 grams of sodium chloride (NaCI). Using dimensional analysis, calculate how many moles of NaCI he we
Neporo4naja [7]

Answer:

0.119 moles

Explanation:

Given that,

Mass measured by Nurse Antonio is 7 grams of NaCl

To find,

The no of moles of NaCl

Solution,

The number of moles is given by

n=\dfrac{m}{M}

m is given mass

M is molar mass

For NaCl, molar mass is 23+35.5 = 58.5 grams

So,

n=\dfrac{7}{58.5}\\\\n=0.119\ \text{moles}

Therefore, there are 0.119 moles of NaCl.

6 0
3 years ago
IF 14.07*10^26 molecules of magnesium chloride was produced in the following reaction, how many grams of magnesium reacted?
Wittaler [7]

Answer:

56160grams

Explanation:

First, we need to convert the number of molecules of magnesium chloride (MgCl2) into moles by dividing by Avagadro's number (6.02 × 10^23 molecules)

n = nA ÷ 6.02 × 10^23

n = 14.07 × 10^26 ÷ 6.02 × 10^23

n = 14.07/6.02 × 10^(26-23)

n = 2.34 × 10^3 moles of MgCl2

The balanced reaction given in the question is as follows:

Mg + 2HCl → MgCl2 + H2

If 1 mole of Mg produced 1 mole of MgCl2

Then, 2.34 × 10^3 moles of Mg will also produce 2.34 × 10^3 moles of MgCl2.

Using mole = mass ÷ molar mass (MM)

Molar mass of Mg = 24g/mol

mass = mole × MM

mass = 2.34 × 10^3 × 24

mass = 56.16 × 10^3

mass = 56160grams.

6 0
3 years ago
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