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r-ruslan [8.4K]
3 years ago
5

Write the steps to separate the Water from salt solution​

Chemistry
2 answers:
zhuklara [117]3 years ago
8 0

Answer:

you can use FILTRATION

STEP :-

1) put a hot water into breaker .

2) stir the solution until the sand dissolve

3) filtrate the solution

4) take the filtrate and evaporate under the sun

then you can get the water and salt too

Hope this helps :)

navik [9.2K]3 years ago
7 0

Answer:

try use distillation

you also can get the salt by evaporate them under the sun

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Number of Protons: 45

Number of Neutrons: 58

Number of Electrons: 45

Explanation:

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What type of rock commonly often contains gas bubbles and the mineral(s) hornblende, muscovite, olivine, lucite, and/or plagiocl
olga_2 [115]

Scoria

Explanation:

Scoria is a  rock that is commonly made up of gas bubbles and minerals such as hornblende, muscovite, olivine, lucite and or plagioclase.

Scoria is an extrusive igneous rock with a lot of cavities and vesicles indicating gas bubbles.

  • Scoria is product of basaltic magma.
  • It has nearly the same composition as basalt since it originates from it.
  • They silica deficient and very fine grained with holes in them.
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  • They are usually found associated with cindercone volcanoes.

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4 0
3 years ago
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
Brums [2.3K]

Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

7 0
3 years ago
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