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3 years ago

How to writhe the balanced chemical equation

Chemistry

1 answer:

mariarad [96]3 years ago

3 0

Get an equal amount of each element by adding coefficients to the different compounds.

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Answer the ones please get right

Natali [406]

Answer:

Hope this helps!

Explanation:

I would help you, but I won't. Because you take others people points without helping them and but random words. So, since you are doing that I won't be helping you <3. Thanks for the points idiot.

6 0

3 years ago

An aqueous solution is listed as being 33.8% solute by mass with a density of 1.15 g/mL, the molar mass of the solute is 145.6 g

vodomira [7]

Answer:

A) 2.69 M

B) 0.059

Explanation:

A) We have:

33.8% solute by mass= 33.8 g solute/100 g solution

molarity = mol solute/ 1 L solution

molarity= $\frac{33.8 g solute}{100 g solution}$ x $\frac{1.15 g solution}{1 ml}$ x $\frac{1 mol solute}{145.6 g solute}$ x $\frac{1000 ml}{1 L}$

molarity= 2.69 mol solute/L solution = 2.69 M

B) We know that there are 33.8 g of solute in 100 g of solution.

As the total solution is compounded by solute+solvent (in this case, solvent is water), the mass of water is the difference between the mass of the total solution and the mass of solute:

mass of water= 100 g - 33.8 g = 66.2 g

Now, we calculate the number of mol of both solute and water:

mol solute= 33.8 g solute x $\frac{1 mol solute}{145.6 g}$ = 0.232 mol

mol H20= 66.2 g H₂O x $\frac{1 mol H2O}{18 g}$

Finally, the mol fraction of solute (Xsolute) is calculated as follows:

Xsolute= $\frac{mol solute}{total mol}= \frac{mol solute}{mol solute + mol H2O}=\frac{0.232 mol}{0.232 mol + 3.677 mol}$

Xsolute= 0.059

4 0

3 years ago

Were you in ms.secrihrst class in dutchtown middle.

IrinaVladis [17]

NO, why though this is a simple question lol!

4 0

3 years ago

Two trials are run, using excess water. In the first trial, 7.8 g of Na2O2(s) (molar mass 78 g/mol) is mixed with 3.2 g of S(s).

Tema [17]

Answer:

The answer is "Option C".

Explanation:

Given equation:

$2Na_20_2 (s)+S(s)+2H_2O \longrightarrow 4NaOH(aq)+SO_2(aq)$

$\to \Delta H^{\circ}_{rxn} (298\ K) = -610 \frac{kJ}{mol}$

$\to Na_2O_2 \ Mass = 7.8 \ g\\\\ \to Na_2O_2 \ Molar \ mass = 78 \frac{g}{mol}$

$Na_2O_2$ Has been the reactant which is limited since the two experiments are equal to $Na_2O_2$ for relationship between stress amounts.

$Na_2O_2, n =\frac{Mass of Na_2O_2}{Molar mass of Na_2 O_2}=\frac{7.8 \ g}{78 \frac{g}{mol}} =0.1 \ mol \\\\q=\Delta H^{\circ}_{rxn} \times n = \frac{ -610 \ kJ}{ 2 \ mol \ Na_2 O_2} \times 0.1 \ mol \ Na_2O_2= 30.5 \ KJ\\\\$

Limiting reactant = $Na_2O_2$

$q=30.5 \ kJ \approx 30 \ kJ$

8 0

3 years ago

Aluminum bromide and chlorine gas react to<br> form aluminum chloride and bromine gas.

Margarita [4]

Answer:

Explanation:

4 0

3 years ago