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Volgvan
3 years ago
7

Consider dissolving sugar in water and magnesium sulfate in water. is any new substance being produced? what happens after sodiu

m hydroxide is added to the dissolved magnesium sulfate? does the addition of sodium hydroxide cause a new substance to appear? if so, what evidence shows this?

Chemistry
1 answer:
Olenka [21]3 years ago
8 0

Answer:

Here's what I get  

Explanation:

1. Sugar

(a) Dissolving in water

The white solid dissolves in water to give a colourless solution. There is no evidence that a new substance is being produced.

(b) Addition of sodium hydroxide

Adding the colourless solution of sodium hydroxide to the colourless sugar solution gives a colourless solution. There is no evidence that a new substance is being produced.

2. Magnesium sulfate

(a) Dissolving in water

The colourless crystals dissolve in water to give a colourless solution. There is no evidence that a new substance is being produced.

(b) Addition of sodium hydroxide

Adding the colourless solution of sodium hydroxide to the colourless solution of magnesium sulfate gives a white precipitate (see image). This is evidence that a new substance is being produced.

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Answer:

30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.

Explanation:

M1 = 3.00 M

M2 = 1.80 M

V2 = 50 .0 mL = 50 /1000 L = 0.05 L

V1 = unknown

In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.

So we use the equation:

                                  M1 V1 = M2 V2

V1 = M2 V2 / M1

V2 = 1.80 * 0.05 / 3.0

V2 = 0.09 /3.0

V2 = 0.03 L or 30 mL

To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.

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4 years ago
What generates electricity
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Answer:

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Explanation:

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3 years ago
What mass of ice can be melted with the same quantity of heat as required to raise the temperature of 3.00 mol H2O(l) by 50.0°C?
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Answer:

m=33.9g

Explanation:

Hello,

In this case, we can first compute the heat required for such temperature increase, considering the molar heat capacity of water (75.38 J/mol°C):

Q=nCp \Delta T=3.00mol*75.38\frac{J}{mol\°C} *50.0\°C\\\\Q=11307J

Afterwards, the mass of ice that can be melted is computed by:

Q=n \Delta _{fus}H

So we solve for moles with the proper units handling:

n=\frac{Q}{\Delta _{fus}H} =\frac{11307J}{6010\frac{J}{mol} } =1.88mol

Finally, with the molar mass of water we compute the mass:

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Best regards.

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