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sammy [17]
2 years ago
15

calculate the (m/v) of nacl of a solution made by diluting 25.0 ml of a 7.50 (m/v)% NaCl solution to a total volume of 75.00 ml

Chemistry
2 answers:
postnew [5]2 years ago
8 0

Answer:

c

Explanation:

Elza [17]2 years ago
7 0

Answer:

2.5%

Explanation:

Given

M1 = 7.50%

M2 = ?

V1 = 25.0mL

V2 = 75.00mL

M1V1=M2V2

7.5x25=M2x75

M2=2.5%

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The answer 
<span>the molar ratio for the following equation 
____C3H8 + ____O2 Imported Asset ____CO2 + ____ H2O

</span>after it has been properly balanced:
__1_C3H8 + ____5O2 Imported Asset ____3CO2 + ____ 4H2O

proof:
number of C =3  (C3H8;   3CO2)
number of H =8 (C3H8 ; 4H2O)
number of O = 10(5x2) or (2x3+4)  (5O2;4H2O)

the answer is 
<span>Reactants: C3H8 = 1, O2 = 8; Products: CO2 = 3 and H2O = 4</span>
4 0
3 years ago
Question 4 of 10 Which phrase describes a scientific law? A. A claim that the volume and temperature of a gas are unrelated B. A
Serjik [45]

Answer:it’s C

Explanation:

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8 0
3 years ago
The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air
ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

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That would be 1 mol reacting to release of ethanol,

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Now,

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Given that :

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m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0
3 years ago
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icang [17]

Answer:

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Explanation:

The whole question can be found in the file attached.

The water vapor inside the air freezes thru the entrance of its nitrogen. This is because liquid nitrogen has a very low temperature and seems to be sufficiently cold to condensed and freeze the steam of water. At air pressure, it has a boiling temperature of -196°C. Freezing the skin producing freeze or cold burns can be associated with direct contact.

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2 years ago
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