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3 years ago

What volume does 49g of KBr need to be dissolved in to make a 8.48% (w/v) solution?

Chemistry

1 answer:

dybincka [34]3 years ago

5 0

Answer:

Volume of solution = 577.83 ml (Approx)

Explanation:

Given:

% of (W/V) = 8.48%

Weight of KBr in gram = 49 gram

Find:

Volume of solution

Computation:

% of (W/V) = Weight of KBr in gram / Volume of solution

8.48% = 48 / Volume of solution

Volume of solution = 49 / 8.48%

Volume of solution = 577.83 ml (Approx)

You might be interested in

How many moles are in 297 g of NH3?

BigorU [14]

Answer:

1. 17.4 moles.

2. 1.13 moles

3. 315.5 moles

4. 390.6g

5. 1.13x10⁶ moles

6. 14.8 moles

7. 337 moles

8. 2.15x10²⁴ molecules

9. 3.13x10²⁴ atoms

10. 1.38x10²⁴ particles

11. 1517g

12. 455g of CaF₂

Explanation:

We can convert formula units to moles or vice versa using Avogadro's number and moles to grams using molar mass of the substance:

1. Molar mass NH3: 17.031g/mol

297g * (1mol / 17.031g) = 17.4 moles

2. Molar mass MgCO3: 84.3g/mol

95g * (1mol / 84.3g) = 1.13 moles

3. Using Avogadro's number (6.022x10²³ formula units / mol):

1.9x10²⁶FU * (1mol / 6.022x10²³FU) = 315.5 moles

4. Molar mass H2O: 18g/mol

21.7mol * (18g / mol) = 390.6g

5. Using Avogadro's number (6.022x10²³ molecules / mol):

6.78x10²⁹molecules * (1mol / 6.022x10²³FU) = 1.13x10⁶ moles

6. 8.9x10²⁴FU * (1mol / 6.022x10²³FU) = 14.8 moles

7. Using Avogadro's number (6.022x10²³ atoms / mol):

2.03x10²⁶atoms* (1mol / 6.022x10²³FU) = 337 moles

8. 3.569mol * (6.022x10²³ molecules / 1mol) = 2.15x10²⁴ molecules

9. 5.2mol * (6.022x10²³ atoms / 1mol) = 3.13x10²⁴ atoms

10. Molar mass Li₂SO₄: 109.94g/mol:

36g * (1mol / 109.94g) * (6.022x10²³ molecules / 1mol) * (7 particles / 1molecule) = 1.38x10²⁴ particles

<em>Assuming particles are atoms and in 1 molecule of Li₂SO₄ you have 7 atoms.</em>

11. Molar mass Cl₂: 70.9g/mol:

21.4mol * (70.9g / mol) = 1517g

12. Molar mass CaF₂: 78.07g/mol:

3.51x10²⁴FU * (1mol / 6.022x10²³FU) * (78.07g / mol) = 455g of CaF₂

8 0

3 years ago

Which of the following atoms forms an ionic bond with a sulfur atom? (Points : 3)

elena-s [515]

Ionic bonds are formed when there is complete transfer of valence electrons between two atoms.

Electronegativity tells the trend of an atom to atract electrons.

You should search for the complete set of rules that indicate whether an ionic or covalent bond happens.

There are two relevant rules to state if whether an ionic bond will happen:

- When the difference of electronegativities between the two atoms is greater than 2.0, then the bond is ionic.

- When the difference is between 1.6 and 2.0, the bond is ionic if one of the elements is a metal.

You need to list the electronegativities of the five elements (there are tables with this information)

Element electronegativity

Cu: 1.9
H: 2.2
Cl 3.16
I: 2.66
S: 2.58

Differences:

Cu / S: 2.58 - 1.9 = 0.68
H / S: 2.58 - 2.2 = 0.38
Cl / S: 3.16 - 2.58 =0.58
I / S: 2.66 - 2.58 = 0.08

Those differences are too low to consider that the bond is ionic.

Then the answer is that none of those atoms forms an ionic bond with sulfur.

8 0

3 years ago

Read 2 more answers

For the equilibrium

Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

$H_2S=0.596M$

$H_2=0.004 M$

$S_2=0.002 M$

Explanation:

We are given that for the equilibrium

$2H_2S\rightleftharpoons 2H_2(g)+S_2(g)$

$k_c=9.0\times 10^{-8}$

Temperature, $T=700^{\circ}C$

Initial concentration of

$H_2S=0.30M$

$H_2=0.30 M$

$S_2=0.150 M$

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles of reactant reduced and form product

Concentration of

$H_2S=0.30+2x$

$H_2=0.30-2x$

$S_2=0.150-x$

At equilibrium

Equilibrium constant

$K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

Substitute the values

$9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}$

By solving we get

$x\approx 0.148$

Now, equilibrium concentration of gases

$H_2S=0.30+2(0.148)=0.596M$

$H_2=0.30-2(0.148)=0.004 M$

$S_2=0.150-0.148=0.002 M$

3 0

3 years ago

A redox reaction can also be a?

AveGali [126]

Losing eltron is the answer

7 0

3 years ago

When can all the coefficients in a balanced equation represent volume ratios? when reactants and products are solids at stp when

Alexxx [7]

Answer: when reactants and products are gases at STP.

Justification:

1) STP stands for standard temperature (0°) and pressure (1 atm).

2) According to the kinetic molecular theory of the gases, and as per Avogadro's principle, equal volumes of gases, at the same temperature and pressure, have the same number of molecules.

3) Since the coefficients in a balanced chemical equation represent number of moles, when reactants and products are gases at the same temperature and pressure, the mole ratios are the same that the volume ratios, and then the coefficients of the chemical equation represent the volume ratios.

8 0

3 years ago

Read 2 more answers