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2 years ago

I need the answer in 15 mins pls urjent

Chemistry

1 answer:

Sedbober [7]2 years ago

7 0

Answer:

I don't know I'm sorry I will tell you another answer asks me to

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Measurements show that the energy of a mixture of gaseous reactants increases by during a certain chemical reaction, which is ca

TEA [102]

<h2>Endothermic and Exothermic</h2>

Explanation:

An exothermic response discharges heat and endothermic absorbs to heat
The whole of the enthalpies of the reactants is more prominent than the product, the response will be exothermic
In the product that the items side has a bigger enthalpy, the reaction is endothermic
An exothermic method is one that emits heat. This heat is transferred to the environment
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2 years ago

3. What is the mass of 5 moles of Hydrogen sulfate?I

solong [7]

Answer:

490

98 for 1 mole, Hence for 5 moles 5 X 98 =490.

Explanation:

Brainliest please?

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2 years ago

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Ierofanga [76]

i dont now but thaks you po por the pionts

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2 years ago

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Are nails in a jar with water a chemical change

Digiron [165]

I would say it is a chemical change because after a while, the nails start to rust. Now they only rust after about a week though. But rust is a chemical change.

7 0

3 years ago

Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t

Liono4ka [1.6K]

Answer: The value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

$\Delta T_{b} = 3.80^{o}C$

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

$Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m$

Now, the values of $K_b$ is calculated as follows.

$\Delta T_{b} = i\times K_{b} \times m$

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

$K_{b}$ = molal boiling point elevation constant

Substitute the values into above formula as follows.

$\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m$

Thus, we can conclude that the value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

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2 years ago