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3 years ago

When seawater freezeswhat happens to the salt?

Chemistry

1 answer:

Studentka2010 [4]3 years ago

8 0

Answer:

Explanation:

pure water can accumulate to form ice while pushing the salt particles to be accumulated in the liquid part, because salt particles are more favorable to stay in liquid than ice

so there can still be some salt in the ice between the bonds

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Sample A: 300 mL of 1M sodium chloride

yarga [219]

Answer:

sample B contains the larger density

Explanation:

Given;

volume of sample A, V = 300 mL = 0.3 L

Molarity of sample A, C = 1 M

volume of sample B, V = 145 mL = 0.145 L

Molarity of sample B, C = 1.5 M

molecular mass of sodium chloride, Nacl = 23 + 35.5 = 58.5 g/mol

Molarity is given as;

$C = \frac{moles \ of \ solute, \ mol}{liters \ of \ solvent} \\\\Moles \ of \ solute \ for \ sample \ A = 1 \times 0.3 = 0.3 \ mol\\\\Moles \ of \ solute \ for \ sample \ B = 1.5 \times 0.145 = 0.2175 \ mol$

The reacting mass for sample A = 0.3mol x 58.5 g/mol = 17.55 g

The reacting mass for sample B = 0.2175 mol x 58.5 g/mol = 12.72 g

The density of sample A $= \frac{mass}{volume} = \frac{17.55}{0.3} = 58.5 \ g/L$

The density of sample B $= \frac{mass}{volume} = \frac{12.72}{0.145} = 87.72 \ g/L$

Therefore, sample B contains the larger density

5 0

3 years ago

A solution with a [H+] of 1.22x10-3M has a pH of _____________________.

Kryger [21]

Answer:

2.91

Explanation:

pH=-log(H3O+)

- Hope that helps! Please let me know if you need further explanation.

7 0

3 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes

frez [133]

Answer : The mass of $MnCO_3$ required are, 35 kg

Explanation :

First we have to calculate the mass of $MnO_2$ .

The first step balanced chemical reaction is:

$2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2$

Molar mass of $MnCO_3$ = 115 g/mole

Molar mass of $MnO_2$ = 87 g/mole

Let the mass of $MnCO_3$ be, 'x' grams.

From the balanced reaction, we conclude that

As, $(2\times 115)g$ of $MnCO_3$ react to give $(2\times 87)g$ of $MnO_2$

So, $xg$ of $MnCO_3$ react to give $\frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg$ of $MnO_2$

And as we are given that the yield produced from the first step is, 65 % that means,

$60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg$

The mass of $MnO_2$ obtained = 0.4542x g

Now we have to calculate the mass of $Mn$ .

The second step balanced chemical reaction is:

$3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3$

Molar mass of $MnO_2$ = 87 g/mole

Molar mass of $Mn$ = 55 g/mole

From the balanced reaction, we conclude that

As, $(3\times 87)g$ of $MnO_2$ react to give $(3\times 55)g$ of $Mn$

So, $0.4542xg$ of $MnO_2$ react to give $\frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg$ of $Mn$

And as we are given that the yield produced from the second step is, 80 % that means,

$80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg$

The mass of $Mn$ obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of $MnCO_3$ required are, 35 kg

4 0

3 years ago

Read 2 more answers

What is the nuclear equation for decay of Am - 241

adelina 88 [10]

Americium is a radioactive element .

It goes through alpha decay during nuclear reactions .

The nuclear reaction.is given by

$\rm {}^{241}_{95}Am\longrightarrow {}^{237}_{93}Np+{}^4_2\alpha$

Np stands for Nuptunium

Here alpha can be written as He too

3 0

1 year ago

Lab activity - Law of Conservation of Mass with vinegar and baking soda.

lesya [120]

Answer: I BIG DUMB

Explanation:

7 0

3 years ago