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Leni [432]
3 years ago
8

Which dilute acid is used to make sodium carbonate salt ​

Chemistry
2 answers:
andreev551 [17]3 years ago
8 0
Hydrochloric acid :))
horsena [70]3 years ago
4 0

Answer:

Sulfuric acid.

The sulfuric acid can react with common salt to produce Na2SO4, and after that you add CaCO3 and carbon to get Na2CO3

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Which molecule contains ten hydrogen atoms
My name is Ann [436]

the answer is hydrocarbons contain 10 hydrogen atoms  :D

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3 years ago
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What is the molarity of a solution that contains 25 g of HCl in 150 mL of solution??
Stels [109]
To find molarity 
1) number of mol of solute.
Solute is HCl.
M(HCl)= 1.0+35.5 =36.5 g/mol

25g *1 mol/36.5 g = 25/36.5 mol HCl

2) Molarity is number of mole of the solute in 1 L solution.
150 mL = 0.150 L

(25/36.5 mol HCl )/(0.150 L) = 25/(36.5*0.150) ≈ 4.57≈4.6 mol/L


3 0
3 years ago
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What determines crystal size in minerals formed by lava or magma?
Burka [1]

Answer:

When magma cools, crystals form because the solution is super-saturated with respect to some minerals. If the magma cools quickly, the crystals do not have much time to form, so they are very small. If the magma cools slowly, then the crystals have enough time to grow and become large.

Explanation:

5 0
3 years ago
Chemical formula for aluminum hydroxide is Al(OH)3 which of the following is true aluminum
frozen [14]
The answer to yo question is ( it has three OH groups). 
4 0
3 years ago
Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 1.30 MJ o
Sunny_sXe [5.5K]

Answer:

97 000 g Na

Explanation:

The absortion (or liberation) of energy in form of heat is expressed by:

q=m*Cp*ΔT

The information we have:

q=1.30MJ= 1.30*10^6 J

ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)

Cp=30.8 J/(K mol Na)

If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.

To do so, we use the molar mass of Na= 22.99g/mol

Cp= \frac{30.8J}{K*mol Na}*\frac{1 mol Na}{22.99 g Na}=1.34\frac{J}{K*g Na}

Now, we are able to solve for m:

m=\frac{1.30*10^6 J}{1.34\frac{J}{K*g Na} *10.0K}=\frac{1.30*10^6J}{13.4\frac{J}{g Na} }  = 9.70*10^4 g Na=97 Kg Na=97 000 g Na

7 0
3 years ago
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