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2 years ago

What change would increase the amount of solid solute able to be dissolved in liquid water?

Chemistry

1 answer:

exis [7]2 years ago

3 0

Bonjour,

increasing temperature.
for many solids dissolved in liquid water, the solubility increases with temperature.

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2 years ago

What products would form if chlorine gas was bubbled through a solution of sodium bromide?

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2 years ago

why was the mass lost when the reaction was done in the normal setup but stayed the same when it was done in the gas collection

german

<h2>Answer : Law of conservation of mass</h2><h3>Explanation :</h3>

The law of conservation of mass states that in any reaction mass is neither created nor lost it has to remain constant in a system.

In this case, when the reaction setup was done in normal way the mass was lost in surrounding was not considered nor being calculated; whereas when the reaction was studied in a closed system where the gas was collected after the reaction the mass changes was noted down which helped to prove the point of law of conservation of mass and energy.

One can consider an example of soda can where the carbonated drink contains pressurized carbon dioxide gas. when opened the gas bubbles gets lost into the surroundings and we don't measure the mass changes. Instead if the soda can was opened in such a way where the gas evolved was measured then the mass changed would remain the same.

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3 years ago

I have to draw a diagram of the ingredient's particles in a pancake. E.g milk eggs etc the diagram must include what the particl

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3 years ago

The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

<u>Answer:</u> The empirical formula for the given compound is $C_3H_6O$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

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3 years ago