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leva [86]
3 years ago
5

What volume would be needed to prepare 375 mL of a .45 M CaCl2 using only a solution of 1.0 M CaCl2 and water?

Chemistry
1 answer:
Neporo4naja [7]3 years ago
8 0

Answer:

168.75 ml

Explanation:

M1V1=M2V2

375ml*.45M=1M*V2

You might be interested in
A solutin contains 100.0 g water, 10.0 g NaCl, and 15.0 g Methanol. What is the Mass Percent of methanol in the solution? The an
Debora [2.8K]
Mass percent of methanol :

% mass = mass methanol / mass total

% mass = 15.0   / 100.0 + 10.0 + 15.0

% mass = 15.0 / 125.0

% mass ( methanol ) = 0,12 x 100 = 12%

hope this helps!

4 0
3 years ago
The mixture of carbon dioxide and water tastes salt why​
vampirchik [111]

Answer:

When carbon dioxide reacts with water carbonic acid is formed, from which hydrogen ions dissociate increasing the acidity of the systemCarbon dioxide emissions to the atmosphere can therefore increase the acidity of land, sea and air

Explanation:

6 0
3 years ago
How many moles are in 124.7 g of Ba(OH)2
Alekssandra [29.7K]

Answer:

0.73 mol

Explanation:

No. of moles(n)= Given mass/molar mass.

Given mass=124.7g

Molar mass of Ba(OH)2= Molar mass of (Barium+2Oxygen+2Hydrogen)=137+32+2=171g

No. of moles= 124.7g/171g=0.73 mol

7 0
3 years ago
What's the formula equation for the reaction that occurred between the barium chloride solution , BaCl2 (aq), and the sodium sul
Solnce55 [7]
The balanced equation is:

BaCl2 (aq) + Na2SO4 (aq) ----> BaSO4(s)+ 2 NaCl(aq)

This is a double replacement reaction.

The reactants are:

a) BaCl2: barium chloride, a ionic compound, therefore soluble in water,

b) Na2SO4: sodium sulfate, another ionic compound, therefore also soluble in water.

The products are:

c) BaSO4: barium sulfate, a solid not soluble in water which precipitates.

d) NaCl: sodium chloride, an ionic compound, therefore soluble in water.
4 0
3 years ago
A sample of hydrogen gas has an initial pressure of 2.86 atm and an initial volume of 8472 mL If the volume of the
Oliga [24]
<h3>Answer:</h3>

Gas law : Boyle's law

New pressure: 66.24 atm

<h3>Explanation:</h3>

Concept tested: Gas laws (Boyle's law)

<u>We are given,</u>

  • Initial pressure, P₁ = 2.86 atm
  • Initial volume, V₁ = 8472 mL
  • New volume, V₂ IS 365.8 mL

We need to determine the new pressure, P₂

  • According to Boyle's law , the volume of a fixed mass of a gas and the pressure are inversely proportional at constant temperature.
  • That is, P\alpha \frac{1}{V}
  • This means , PV = k (constant)
  • Therefore; P₁V₁ = P₂V₂
  • Rearranging the formula, we can get the new pressure, P₂

P₂ = P₁V₁ ÷ V₂

   = (2.86 atm × 8472 mL) ÷ 365.8 mL

   = 66.24 atm

Therefore, the new pressure is 66.24 atm

5 0
3 years ago
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