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3 years ago

1. Calcium chloride is a salt that is added to roads and sidewalks

Chemistry

1 answer:

Sholpan [36]3 years ago

8 0

Answer:

(d) In the presence of sulfuric acid, calcium chloride reacts to form hydrogen chloride.

Explanation:

It's because it lowers the melting point of the ice / etc so I think it is, but check it yourself too!

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What Group is this element in?<br> Please answer （＾∇＾）

Aneli [31]

Group 2 because the outer electrons ( also known as valence electrons) are 2 and that’s how we know which group it is in

5 0

3 years ago

How to prepared sodium chloride solution.

nexus9112 [7]

Explanation:

Dissolve 93.52g of NaCl in about 400mL of distilled water, then add more water until final volume is 800mL. If starting with a solution or liquid reagent: When diluting more concentrated solutions, decide what volume(V2) and molarity (M2) the final soluble should be.

4 0

3 years ago

1) 0.5 moles of sodium chloride is dissolved to make 0.05 liters of solution. 2) 734 grams of lithium sulfate are dissolved to m

OLEGan [10]

Answer:

1)Molarity of 0.5 Moles of Sodium Chloride in 0.05 Liters of solution is 10 M.

2)Molarity of 734 grams of lithium sulfate are dissolved to make 2500 mL of solution is 2.68 M.

3)Molarity of a solution is made by adding 83 grams of sodium hydroxide to 750 mL of water is 2.77 M.

4)Molarity of a solution that contains .500 mol HC₂H₃O₂ in 0.125 kg H₂O is 4 M.

5)Molarity of a solution that contains 63.0 g HNO₃ in 0.500 kg H₂O is 2 M.

6)The mass of water required to form 3.0 M solution by dissolving 0.5kg of C₂H₅OH is 3.62 kg.

Explanation:

Molarity is given as

$M=\dfrac{n}{V}$

Here

n is number of moles
V is the volume of the solution in liters

Using this all the values are calculated as follows:

1)Molarity of 0.5 Moles of Sodium Chloride in 0.05 Liters of solution is

$M=\dfrac{n}{V}\\M=\dfrac{0.5}{0.05}\\M=10$

So the molarity is 10 M.

2)Molarity of 734 grams of lithium sulfate are dissolved to make 2500 mL of solution is

$M=\dfrac{n}{V}$

Here n is calculated as follows:

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}$

Here the molar mass of lithium sulphate is as follows:

$MM of Li_2SO_4=2\times Li+S+4\times O\\MM of Li_2SO_4=2\times 7+32+4\times 16\\MM of Li_2SO_4=110$

So n is

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}\\n=\dfrac{734}{110}\\n=6.67$

V is given as

$V_{L}=\dfrac{V_{mL}}{1000}\\V_{L}=\dfrac{2500}{1000}\\V_{L}=2.5\ L$

So the molarity is given as

$M=\dfrac{n}{V}\\M=\dfrac{6.67}{2.5}\\M=2.68\ M$

So the molarity is 2.68 M.

3)Molarity of a solution is made by adding 83 grams of sodium hydroxide to 750 mL of water is

$M=\dfrac{n}{V}$

Here n is calculated as follows:

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}$

Here the molar mass of Sodium hydroxide is as follows:

$MM of NaOH=Na+O+H\\MM of NaOH=23+16+1\\MM of NaOH=40$

So n is

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}\\n=\dfrac{83}{40}\\n=2.075$

V is given as

$V_{L}=\dfrac{V_{mL}}{1000}\\V_{L}=\dfrac{750}{1000}\\V_{L}=0.75\ L$

So the molarity is given as

$M=\dfrac{n}{V}\\M=\dfrac{2.075}{0.75}\\M=2.77\ M$

So the molarity is 2.77 M.

4)Molarity of a solution that contains .500 mol HC₂H₃O₂ in 0.125 kg H₂O is

$M=\dfrac{n}{V}$

V is given as

$V_{L}=\dfrac{mass}{density}\\V_{L}=\dfrac{0.125}{1 kg/L}\\V_{L}=0.125\ L$

So the molarity is given as

$M=\dfrac{n}{V}\\M=\dfrac{0.5}{0.125}\\M=4.0\ M$

So the molarity is 4.00 M.

5)Molarity of a solution that contains 63.0 g HNO₃ in 0.500 kg H₂O is

$M=\dfrac{n}{V}$

Here n is calculated as follows:

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}$

Here the molar mass of HNO₃ is as follows:

$MM of HNO_3=H+N+3\times O\\MM of HNO_3=1+14+3\times 16\\MM of HNO_3=63$

So n is

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}\\n=\dfrac{63}{63}\\n=1.00$

V is given as

$V_{L}=\dfrac{mass}{density}\\V_{L}=\dfrac{0.5}{1 kg/L}\\V_{L}=0.5\ L$

So the molarity is given as

$M=\dfrac{n}{V}\\M=\dfrac{1}{0.5}\\M=2.00\ M$

So the molarity is 2.00 M.

6)Mass of water must be used to dissolve 0.500 kg C₂H₅OH to prepare a 3.00 m solution is calculated as follows

$M=\dfrac{n}{V}$

Here n is calculated as follows:

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}$

Here the molar mass of C₂H₅OH is as follows:

$MM of C_2H_5OH=2\times C+6\times H+O\\MM of C_2H_5OH=2\times 12+6\times 1+16\\MM of C_2H_5OH=46$

So n is

$n=\dfrac{mass\ of\ solute}{Molar\ mass\ of\ Solute}\\n=\dfrac{500}{46}\\n=10.87$

V required is given as

$M=\dfrac{n}{V}\\V=\dfrac{n}{M}\\V=\dfrac{10.87}{3}\\V=3.62\ L$

So the mass of water is given as

$mass=density\times Volume\\mass=1 kg/L \times 3.62 L\\mass=3.62\ kg$

So the mass of water required to form 3.0 M solution by dissolving 0.5kg of C₂H₅OH is 3.62 kg.

6 0

2 years ago

Would chromium-51 be useful for dating rocks containing chromium?

Over [174]

The half-life of chromium-51 is a very short period of time so no because it would be so little left that it would be difficult to detect.
Hope this helps!!

8 0

2 years ago

Many laboratory gases are sold in steel cylinders with a volume of 43.8 L. What mass (in grams) of argon is inside a cylinder wh

MrRa [10]

It's very simple... if we remember value of Universal Gas Constant R and Ideal Gas Law, so...

Ideal Gas Law
pV = nRT, where:
p - pressure (in kPa),
V - volume (in L),
n - number of moles (in mol),
R - universal cas constant (in kPa * L / mo l* K),
T - temperature (in K)

n = m/M, where:
n - number of moles,
m - mass (in grams),
M - molar mass of ingredient (in g/mol) - you find this at Periodic Table.

pV = nRT ---> pV = mRT/M ---> pVM = mRT ---> pVM/RT = m

p = 17615 kPa
T = 273.15 + 23 = 296.15 K
V = 43.8 L
R = 8.314 kPa * L / mol * K
M (for argon) = 39.948 g/mol

and

m = (17615 kPa * 48.3 L * 39.948 g/mol) / (296.15 K * 8.314 kPa * L / mol * K)
m = 13803.93 grams of Argon

5 0

3 years ago