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serious [3.7K]
4 years ago
5

Help with as many as possible is appreciated! :)

Chemistry
1 answer:
pav-90 [236]4 years ago
4 0

Answer:

a. 0.0855 M

b. 0.153 M

c. 0.0351 M

d. 0.0901 M

Explanation:

a.

Molar mass NaCL= 23.0 +35.5 = 58.5 g/mol

10.0 g* 1 mol/58.5 g = 10.0/58.5 mol

(10.0/58.5) mol/2L = 0.0855 mol/L = 0.0855 M

b.

Molar mass (C12H22O11) =342.3 g/mol

52.5 g*1mol/342.3 g = 52.5/342.3 mol=0.153 mol

0.153 mol/1L = 0.153 M

c.

Molar mass (Al2(SO4)3) =342.2 g/mol

120g* 1 mol/342.2 g =0.351 mol

0.351 mol/10.0L = 0.0351 mol/L = 0.0351 M

d.

Molar mass (C8H10N4O2) = 194.2 g/mol

1.75 g*1mol/194.2 g = 0.00901 mol

0.00901 mol/0.100 L =0.0901 M

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The value of h for the reaction below is -73 kj how many kj of heat are released
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If the value of H is positive, it means you have to add that much heat to complete the reaction. If H is negative, it means that much heat is released during the chemical process. Because it is -73 kJ, 73 kJ of heat are released in the reaction.
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3 years ago
I think it’s D but I’m not sure.. Please help!
alisha [4.7K]

Answer:   I think it’s D?

Explanation:

7 0
3 years ago
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8 points
butalik [34]

Answer:

1.3 meters

Explanation: use newton third law equation.

5 0
3 years ago
Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50
devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C_{V} You can work out C_{V}

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC_{V} ∆T

Polyatomic gas: C_{V} = 3R

∆U= nC_{V} ∆T

∆U= 28g x C_{V} x (350K - 58.3K)

∆U = 28C_{V} x 291.7

∆U = 10967.6 x C_{V}

5 0
3 years ago
Pure substances are _______ in appearance ​
harina [27]

Answer:

constant appearance

Explanation:

this mean that a pure substance will have a constant appearance ,colour,density ,melting point and boiling point

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3 years ago
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