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Lady bird [3.3K]
2 years ago
15

Hello! Please help.

Chemistry
1 answer:
loris [4]2 years ago
4 0

Answer:

Air is a mixture. Its constituents can be separated. For example: oxygen, nitrogen etc.

Hydrochloric acid is a mixture, being an acidic liquid.

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What holds all the planets and objects in the solar system in place?
uysha [10]

Answer:

Gravity

Explanation:

i dont rlly know where it comes from...

5 0
3 years ago
Can someone help me? It needs to have a diagram that has arrows.
daser333 [38]

Answer: The enthalpy change for formation of butane is -125 kJ/mol

Explanation:

The balanced chemical reaction is,

C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)

The expression for enthalpy change is,

\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]

Putting the values we get :

\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]

-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]

H_f_{C_4H_{10}=-125kJ/mol

Thus enthalpy change for formation of butane is -125 kJ/mol

5 0
3 years ago
Correct?
erica [24]
Correct answer is C

Na2O + H2O ----> 2 NaOH

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4 0
3 years ago
What features of this model will help Armando answer the question?
ASHA 777 [7]

Answer:

Explanation:

Answer to Armando is an artist who sells prints of his original paintings. ... This problem has been solved! ... Armando wants to create a function P(x). to model his total profit where x is the number of paintings sold

5 0
2 years ago
A 13.1-g sample of CaCl2 is dissolved in 104 g of water, with both substances at 24.7°C. Calculate the final temperature of the
likoan [24]

Answer:

The final temperature of the solution is 44.8 °C

Explanation:

assuming no heat loss to the surroundings, all the heat of solution (due to the dissolving process) is absorbed by the same solution and therefore:

Q dis + Q sol = 0

Using tables , can be found that the heat of solution of CaCl2 at 25°C (≈24.7 °C)  is q dis= -83.3 KJ/mol . And the molecular weight is

M = 1*40 g/mol + 2* 35.45 g/mol = 110.9 g/mol

Q dis = q dis * n  = q dis * m/M =  -83.3 KJ/mol * 13.1 g/110.9 gr/mol = -9.84 KJ

Qdis= -9.84 KJ

Also Qsol = ms * Cs * (T - Ti)

therefore

ms * Cs * (T - Ti) + Qdis = 0

T=  Ti - Qdis * (ms * Cs )^-1   =24.7 °C - (-9.84 KJ/mol)/[(104 g + 13.1 g)* 4.18 J/g°C] *1000 J/KJ

T= 44.8 °C

7 0
3 years ago
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