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Softa [21]
2 years ago
12

We can separate sand from water by using a cloth or filter paper , but we .cannot separate salt from water by the same method.

Chemistry
2 answers:
Nataly_w [17]2 years ago
6 0
Yeh yeh


Yeh sepsreat water lawl
Karolina [17]2 years ago
4 0

the size of the salt is smaller than the pores of the cloth or filter paper. to separate the two we can use evaporation

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Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make ___ mol ammonium sulfate
Bingel [31]

Answer:

1) Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make 0.157 mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.

2) 2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.

Explanation:

  • Firstly, we should balance the equation of heptane combustion.
  • We can balance the equation by applying the conservation of mass to the equation.
  • The balanced equation is: <em>2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.</em>
  • This means that every 2.0 moles of ammonium hydroxide (NH₄OH) will produce 1.0 mole of ammonium sulfate (NH₄)₂SO₄ when it is neutralized by sulfuric acid.
  • We need to calculate the no. of moles in 11.0 g of ammonium hydroxide that is neutralized using the relation: <em>n = mass/molar mass. </em>

n of 11.0 g of ammonium hydroxide (NH₄OH) = mass/molar mass = (11.0 g)/(35.04 g/mol) = 0.314 mol.

<u><em>Using cross multiplication: </em></u>

2.0 moles of NH₄OH make → 1.0 mole of (NH₄)₂SO₄.

0.314 mol of NH₄OH make → ??? moles of (NH₄)₂SO₄.

∴ The no. of moles of (NH₄)₂SO₄ that will be made from neutralizing (11.0 g) of NH₄OH = (0.314 mol)(1.0 mol)/(2.0 mol) = 0.157 mol.

<em>∴ Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make </em><em>0.157</em><em> mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.</em>

3 0
3 years ago
Would the amount of solid magnesium increase decrease or stay the same if you added a strong acid to it?
Vlada [557]
Solid Magnesium is considered as active metal so it reacts with strong acids like HCl and H₂SO₄ liberating Hydrogen gas according to the following equations:

Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g)
Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g)

so the amount of solid magnesium decrease  by addition of strong acid to it.
8 0
3 years ago
What is the surface tension of H20
andrew11 [14]

Answer:

the surface tension of H20 is 72 dynes/cm at 25°C

3 0
3 years ago
Read 2 more answers
2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea
Alex_Xolod [135]

<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.0275 M

<u>Explanation:</u>

Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}

Moles of HI = 0.550 moles

Volume of container = 2.00 L

\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M

For the given chemical equation:

                          2HI(g)\rightleftharpoons H_2(g)+I_2(g)

<u>Initial:</u>                  0.275

<u>At eqllm:</u>           0.275-2x      x         x

The expression of K_c for above equation follows:

K_c=\frac{[H_2][I_2]}{[HI]^2}

We are given:

K_c=0.0156

Putting values in above expression, we get:

0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

4 0
3 years ago
You are running a lemonade stand with your friend. You prepared 10 liters of 0.7 molarity lemonade, but your friend did online r
olga nikolaevna [1]

Answer:

add 7.5L of water

Explanation:

M1×V1=M2×V2

M is molarity, V is volume

0.7 × 10 = 0.4 × V2

V2= 17.5L

vol. of water to add= 17.5 - 10 = 7.5L

3 0
3 years ago
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