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shepuryov [24]
2 years ago
14

at standard pressure, what is the temperature at which a saturated solution of NH4Cl has a concentration of 60g NH4CL/100 g H2O

Chemistry
1 answer:
Nataliya [291]2 years ago
5 0

Answer: Temperature = T, unknown

Saturated Solution, NH4Cl concentration = 60g/100g H2O = 0.6g NH4Cl/g H2O

Assume density of H2O = 1 g/ml

m = 0.6g NH4Cl/g H2O / 1 g/ml

m = 0.6g NH4Cl/ml

See the table of saturated solutions and identify the temperature at which the concentration of NH4Cl is 60g/100g H2O.

Explanation:  The line on the graph on reference table G indicates a saturated solution of NH4CL as a concentration of 60. g NH4 Cl/100. g H2O

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ArbitrLikvidat [17]

Answer:

c. 3.00 M HCl

Explanation:

From dilution formula

C1V1 = C2V2

C1=?, V1= 10.0ml, C= 1.5, V2= 20.0ml

Substitute and Simplify

C1×10= 1.5×20

C1= 3.00M

5 0
3 years ago
When all colors are absorbed, you see:<br><br> black<br> white<br> green<br> yellow
wolverine [178]

uhmm, white.

Explanation:

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3 years ago
Ionic equation for reaction of koh(aq) with cucl2(aq)
marishachu [46]

Answer:

2 KOH(aq) + CuCl2(aq) = 2 KCl(s) + Cu(OH)2

Explanation:

8 0
3 years ago
Solution A is 0.44 M and reacts with 0.11 M of solution B. Assume that the value of x is 0, the value of y is 1, and r is 1.07 ×
xz_007 [3.2K]

Answer:

K, the rate constant = 9.73 × 10^(-1)/s

Explanation:

r = K × [A]^x × [B]^y

r = Rate = 1.07 × 10^(-1)/s

K = Rate constant

A and B = Concentration in mol/dm^-3

A = 0.44M

B = 0.11M

x = Order of reaction with respect to A = 0

y = Order of reaction with respect to B = 1

Solving, we get

r/([A]^x × [B]^y) = K

K = 1.07 × 10^(-1)/s/(0.44^0 × 0.11^1)= 0.9727

K = 0.9727

7 0
3 years ago
What is the molar mass of 37.96 g of gas exerting a pressure of 3.29 on the walls of a 4.60 L container at 375 K?
Phantasy [73]

The molar mass of the gas is 77.20 gm/mole.

Explanation:

The data given is:

P = 3.29 atm,   V= 4.60 L   T= 375 K  mass of the gas = 37.96 grams

Using the ideal Gas Law will give the number of moles of the gas. The formula is

PV= nRT    (where R = Universal Gas Constant 0.08206 L.atm/ K mole

Also number of moles is not given so applying the formula

n= mass ÷ molar mass of one mole of the gas.

n = m ÷ x   ( x  molar mass) ( m mass given)

Now putting the values in Ideal Gas Law equation

PV = m ÷ x RT

3.29 × 4.60 = 37.96/x × 0.08206 × 375

15.134 = 1168.1241  ÷ x

15.134x = 1168.1241

x = 1168.1241 ÷ 15.13

x = 77.20 gm/mol

If all the units in the formula are put will get cancel only grams/mole will be there. Molecular weight is given by gm/mole.

8 0
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