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guapka [62]
3 years ago
7

A sample of seawater consists of water, sodium ions, chloride ions, and several other dissolved salts. These ions and salts are

evenly distributed throughout the water.Which term or terms could be used to describe this sample of seawater?Check all that apply.1. mixture2. heterogeneous mixture3. homogenous mixture4. solution5. pure chemical substance6. compound7. element
Chemistry
1 answer:
stellarik [79]3 years ago
6 0

Answer:

Mixture, homogeneous mixture, solution

Explanation:

A mixture is defined as a substance which is made by physical combination of two or more substance.

As in sea water, more than two substance are present, therefore is considered as mixture.

A mixture is considered as homogeneous mixture, if it has same proportion of components through out the mixture. In sea water proportions of all the components are same, therefore, its is also a homogeneous mixture.

In solution, solute is dissolved in a solvent and the concentration of the solute is same through out the solution.. In sea water, sodium ions, chloride ions, and several other salts are dissolved in water. Therefore, water is considered as solvents and components present are considered as solute.

so, sea water can also considered as solution.

Among the given options, 1, 3 and 4 are correct

So, sea water is considered as mixture, homogeneous mixture and solution.

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A 30.2 mL aliquot of a 30.0 wt% aqueous KOH solution is diluted to 1.20 L to produce a 0.173 M KOH solution. Calculate the densi
alexandr1967 [171]

Answer:

The density of solution is 1.283 g/mL.

Explanation:

Molarity of the KOH before dilution = M_1

Volume of the solution before dilution = V_1=30.2 mL

Molarity of the KOH after dilution = M_2=0.173 M

Volume of the solution after dilution = V_2=1.20 L=1200 mL

M_1V_1=M_2V_2

M_1=\frac{M_2V_2}{V_1}=\frac{0.173 M\times 1200 mL}{30.2 mL}

M_1=6.8742 M

Molarity=\frac{moles}{Volume (L)}

V_1=30.2 mL=0.0302 L (1 mL = 0.001 L)

M_1=\frac{n}{V_1}

n=M_1\times V_1=6.8742 M\times 0.0302 L=0.2076 mol

Mass of 0.2076 moles of KOH:

0.2076 mol × 56 g/mol = 11.6256 g

Mass of KOH is solution = 11.6265 g

Mass of the solution = M

Mass percentage of solution = 30.0% of KOH

30.0\%=\frac{11.6265 g}{M}\times 100

M = 38.755 g

Density of the solution , d= \frac{M}{V_1}

d=\frac{38.755 g}{30.2 mL}=1.283 g/mL

The density of solution is 1.283 g/mL.

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What’s this answer help me out
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Sorry my guy no idea try googling
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Silver chloride, often used in silver plating, contains 75.27% Ag.
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Answer:

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Explanation:

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How many orbitals are completely filled
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The chemical compound, "AB" has a molar mass of 74.548 g/mol.
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Answer:

47.55%

Explanation:

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