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Nadusha1986 [10]
3 years ago
6

How many grams of solute are needed to prepare a 3.50% mass/mass solution that has a solution mass of 2.50x102 grams.

Chemistry
1 answer:
Elis [28]3 years ago
5 0
One way of expressing concentration is by percent. It may be on the basis of mass, mole or volume. Percent is expressed as the amount of solute per amount of the solution. For this case, we are given the percent by mass. In order to solve the amount of solute, we multiply the percent with the amount of the solution.

Mass of solute = percent by mass x mass solution
Mass of solute = 0.0350 x 2.50 x10^2 = 8.75 grams of solute
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Hi, can someone help me balance this chemistry equation:<br> H2SO4 + RbOH -&gt; Rb2SO4 + H2O
zloy xaker [14]

H2SO4 + 2RbOH -> Rb2SO4 + 2H2O

If you want an explanation, keep reading.

In the first portion, there are two hydrogen ions and four sulfate ions.

The second portion has one rubidium ions and one hydroxide ion.

On the other side of the equation, in order to keep those two rubidiums balanced, you'll need to add a two at the beginning of the second portion, but in that process you are giving a second hydroxide value.

Back to the right side, there is there is water (H2O).

On the first portion, there were two hydrogen ions. The second portion also has two hydroxides because of the value change (adding the two to the front).

So on the fourth portion, you'd have to add another two so you could balance the four hydrogen ions (H2 and 2OH) and the two oxygen ions (2OH).

I hope this was easy to understand.

6 0
3 years ago
Can an ether form hydrogen bonds in the same way as does an ester
belka [17]

Yes, an OH group from ethanol can form a hydrogen bond to the ether O atom in the same way as it can do so with the single-bonded O atom in the ester.

The O atom in the carbonyl group of the ester can also form H-bonds with ethanol.


5 0
3 years ago
How many moles of water are produced when 3.45 moles of KMnO4 react? Give your answer to the nearest 0.1 moles. moles H2O
andrew11 [14]

Answer:

13.8 moles of water produced.

Explanation:

Given data:

Moles of KMnO₄ = 3.45 mol

Moles of water = ?

Solution:

Chemical equation:

16HCl + 2KMnO₄ → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O

Moler ratio of water and KMnO₄:

             KMnO₄       :        H₂O

                2             :          8

               3.45         :         8/2×3.45 = 13.8 mol

Hence, 3.45 moles of KMnO₄ will produced 13.8 mol of water.

7 0
3 years ago
Nickel and carbon monoxide react to form nickel carbonyl, like this: (s)(g)(g) At a certain temperature, a chemist finds that a
horsena [70]

The question is incomplete, here is the complete question:

Nickel and carbon monoxide react to form nickel carbonyl, like this:

Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)

At a certain temperature, a chemist finds that a 2.6 L reaction vessel containing a mixture of nickel, carbon monoxide, and nickel carbonyl at equilibrium has the following composition:

Compound            Amount

     Ni                        12.7 g

   CO                        1.98 g

Ni(CO)_4                  0.597 g

Calculate the value of the equilibrium constant.

<u>Answer:</u> The value of equilibrium constant for the reaction is 2448.1

<u>Explanation:</u>

We are given:

Mass of nickel = 12.7 g

Mass of CO = 1.98 g

Mass of Ni(CO)_4 = 0.597 g

Volume of container = 2.6 L

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Given mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

\text{Equilibrium concentration of nickel}=\frac{12.7}{58.7\times 2.6}=0.083M

\text{Equilibrium concentration of CO}=\frac{1.98}{28\times 2.6}=0.0272M

\text{Equilibrium concentration of }Ni(CO)_4=\frac{0.597}{170.73\times 2.6}=0.00134M

For the given chemical reaction:

Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)

The expression of equilibrium constant for the reaction:

K_{eq}=\frac{[Ni(CO)_4]}{[CO]^4}

Concentrations of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

K_{eq}=\frac{0.00134}{(0.0272)^4}\\\\K_{eq}=2448.1

Hence, the value of equilibrium constant for the reaction is 2448.1

4 0
3 years ago
Name a physical quantity that varies in uniform motion​
nekit [7.7K]

Answer:

the direction of the motion us constantly changing during motion over any close path, not only circular

4 0
2 years ago
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