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Flura [38]
3 years ago
8

When 2.00 g of NaCl are added to 113 g of water in a constant pressure calorimeter, the temperature of the solution increases by

3.08 degrees C. Given that the molar mass of NaCl is 58.44 g/mol, and the specific heat of the solution is 4.184 J/g°C, what is the molar enthalpy of solution, DeltaH solution for NaCl in kJ/mol? Do not report units in your answer. Do use proper significant figures and report your answer using the correct sign.
Chemistry
1 answer:
nata0808 [166]3 years ago
5 0

Answer:

24qe2q43rerret

Explanation:

htgfjhgjdrfadw2q42r4q2qe2wadw2q42r4q2qe2wadw2q42r4q2qe2wadw2q42r4q2qe2wadw2q42r4q2qe2wadw2q42r4q2qe2wadw2q42r4q2qe2wadw2q42r4q2qe2w

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NemiM [27]

Chemistry. More specifically, analytical chemistry.

6 0
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the volume of gas in a container was originally 3.24L, while at standard pressure 1.00 atm. what will the volume be if the press
Simora [160]
Hello!

To solve this problem, we will use the Boyle's Law, which describes how pressure changes when volume changes and vice-versa. The equation for this law is the following one, and we'll clear for V2:

P1*V1=P2*V2 \\ \\ V2= \frac{P1*V1}{P2}= \frac{3,24 L * 1 atm}{1,20 atm}= 2,7 L

So, the final volume after increasing the pressure would be 2,7 L. That means that volume decreases when the pressure increases

Have a nice day!
8 0
3 years ago
14 Copper(II) oxide reacts with hydrogen.
e-lub [12.9K]

Answer:

B

CuO

H2

Explanation:

hydrogen will oxidized to form water

6 0
3 years ago
Which statement is completely accurate?
AURORKA [14]

Answer:The same chemical element can have a different number of neutrons and still be the same element. We refer to the atoms of the same element with different numbers of neutrons as "isotopes"

Explanation:

3 0
2 years ago
At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106
lana [24]

Answer:

The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>          

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M        

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant (K_{c}) for the given chemical reaction, is given by the equation:

K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}

<u><em>At the original equilibrium state:</em></u>

K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}

K_{c} = \frac {1.6641}{0.002332} = 713.59

<u><em>Therefore, at the new equilibrium state:</em></u>

K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}

\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}

\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88

\Rightarrow [HI] = \sqrt {12.88} = 3.589 M

<u>Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M</u>

6 0
2 years ago
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