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4 years ago

11

Give an example, other than making ice-cream or a slushy, where the effect of salt on the phase change of water is useful in eve

ryday life.
ASAP PLS AND THANK YOU

2 answers:

guapka [62]4 years ago

7 0

Answer:

Putting salt on the ground to melt ice or snow I mean.

Explanation:

horrorfan [7]4 years ago

7 0

Using salt to melt sleet and snow not black ice tho u cant melt that

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What salt is formed during the neutralization reaction between sulfuric acid left-parenthesis H subscript 2 S O subscript 4 righ

ivolga24 [154]

Answer:

Li₂SO₄ is the formed salt

Explanation:

We determine the reactants dissociations:

H₂SO₄ → 2H⁺ + SO₄⁻²

LiOH → Li⁺ + OH⁻

The 2H⁺ link the OH⁻ to form water, but there is one more H⁺, so we need another OH⁻ to finally produce 2 moles of H₂O

In conclussion to determine the lithium sulfate, which is the formed salt, Li will release 2 e⁻ as this equation shows:

2Li → 2Li⁺ + 2e⁻

Now the neutralization reaction is:

2LiOH + H₂SO₄ → Li₂SO₄ + 2H₂O

3 0

3 years ago

Write a complete, balanced equation for the following reactions:

lys-0071 [83]

Answer:

Solution given:

c) Magnesium metal plus silver acetate (teachers note:

<u>Balanced</u><u> </u><u>chemical</u><u> </u><u>equation</u><u>:</u>

$\boxed{\bold{\green{Mg+AgC_{2}H_{3}O_{2}\rightarrow MgC_{3}H_{3}O_{2}+Ag^{+}}}}$

d) hydrobromic acid plus zinc metal (teachers note:

<u>Balanced chemical equation:</u>

$\boxed{\bold{\green{2HBr+Zn \rightarrow Be_{2}+H_{2}}}}$

6 0

3 years ago

Millikan's oil-drop experiment A. established the charge on an electron. B. showed that all oil drops carried the same charge. C

kumpel [21]

Answer:

A

Explanation:

This popular experiment enabled us to know the charge of an electron. It was performed by Robert Millikan and Harvey Fletcher and have the process and procedure to measure the elementary electronic charge. In fact, Robert Millikan was awarded the Nobel prize in physics in the year 1923 for how efforts.

They were able to determine the charge by repeating the experiment for several droplets and later proposed that the charges were integer multiples of a particular base value.

8 0

3 years ago

If the solubility of a gas in water is 1.22g/2.75 atm, what is it’s solubility (in g/L) at 1.0 atm

Slav-nsk [51]

Answer : The solubility of a gas in water at 1 atm pressure is, 0.4436 g/L

Solution : Given,

Solubility of a gas in water = 1.22 g/L (at 2.75 atm pressure)

At pressure = 1 atm, Solubility of a gas = ?

Now we have to calculate the solubility of a gas.

At 2.75 atm pressure, the solubility of a gas in water = 1.22 g/L

At 1 atm pressure, the solubility of a gas in water = $\frac{1atm}{2.75atm}\times 1.22g/L=0.4436g/L$

Therefore, the solubility of a gas in water at 1 atm pressure is, 0.4436 g/L

5 0

4 years ago

Which has the greater density: 100 kg of ice or 10 kg of gold? Explain why.

ss7ja [257]

10kg of gold is more denser than 100 kg of ice .

<u>Explanation: </u>

Density of a substance purely depends upon the mass and its volume occupied in the system.

In the above statement it is a comparison made between 100kg of ice vs 10kg of gold.

Density of ice in general is 0.920 $\frac{g}{c m^{3}}$ and Density of gold in general is 19.3 $\frac{g}{c m^{3}}$

So if we calculate for 100kg of ice,

Density=100×0.920=92 $\frac{kg}{c m^{3}}$

Similarly we can calculate for 10kg of gold we get

Density=10×19.13=191.3 $\frac{kg}{c m^{3}}$

If we compare the above two values we can say that

Density of 10kg gold has greater value than the Density of 100kg water.

7 0

4 years ago