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4 years ago

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1 answer:

Illusion [34]4 years ago

8 0

Answer:

A divine the molarity by the volume inliters to find moles

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A cylinder and piston assembly (defined as the system) is warmed by an external flame. The contents of the cylinder expand, doin

LiRa [457]

Answer:

The energy change for the given system: <u>ΔE = 45 J</u>

Explanation:

According to the<u> First Law of thermodynamics</u>, the energy can't be created or destroyed and can only be converted into other forms.

ΔE = Q - W ......equation 1

Given: ΔE is the change in the internal energy

Q is the heat absorbed by the system

W is the work done by the system

<u>Given</u>: Q = = 533 J, W = 488 J, ΔE = ?

Using equation 1, to calculate the change is the internal energy (ΔE),

ΔE = Q - W = 533 J - 488 J = 45 J

<u>Therefore, the energy change for the given system: ΔE = 45 J</u>

4 0

3 years ago

How many grams of LiCl are required to make 2.0 L of 0.65 M LiCl solution?

makkiz [27]

Answer:- 55 g of LiCl are required to make 2.0 L of 0.65 M solution.

Solution:- Given-

molarity of the solution = 0.65M

Volume of solution = 2.0 L

molarity of solution = $\frac{moles of solute}{volume of solution in liter}$

It could be arranged as...

moles of solute = molarity x volume of solution

moles of solute = 0.65 x 2.0 = 1.3 moles

Molar mass of LiCl is 42.39 g/mol.

So, mass of LiCl required = 1.3 mol x 42.39g/mol = 55 g

4 0

3 years ago

What color and texture are best r reflectors of radian heat

lubasha [3.4K]

Answer:

White colors and white surfaces because white surfaces are poor absorbers and good reflectors of heat radiation.

Explanation:

<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>!</em><em> </em>❤️

8 0

3 years ago

Calculate ΔH∘f (in kilojoules per mole) for benzene, C6H6, from the following data: 2C6H6(l) + 15O2(g)→12CO2(g)+6H2O(l) ΔH∘=−653

Pachacha [2.7K]

Answer: 48.6 kJ/mol

Explanation:

The balanced chemical reaction is,

$2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

$\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_6H_6}\times \Delta H_{C_6H_6})]$

where,

n = number of moles

$\Delta H_{O_2}=0$ (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

$-6534.0=[(12\times -393.5)+(6\times -285.8)]-[(15\times 0)+(2\times \Delta H_{C_6H_6})]$

$\Delta H_{C_6H_6}=48.6kJ/mol$

Therefore, the enthalpy change for benzene is 48.6 kJ/mol

4 0

3 years ago

Question 2 of 10

Ganezh [65]

I think A. Molecules interact without bonds breaking

8 0

2 years ago