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ExtremeBDS [4]
3 years ago
12

Help plz:)))I’ll mark u Brainliest

Chemistry
1 answer:
emmasim [6.3K]3 years ago
8 0

Answer:

\Delta _{fus}H=205J/g=13.03kJ/mol

Explanation:

Hello there!

In this case, since the heat of fusion is a property that allows us to calculate the heat involved during the change from solid to liquid (fusion) and is calculated as shown below:

Q=m*\Delta _{fus}H

In such a way, given the heat involved during this process and the mass of copper, we calculate the heat of fusion as shown below:

\Delta _{fus}H=\frac{Q}{m}=\frac{41000J}{200.g}\\\\\Delta _{fus}H=205J/g

Or in kJ/mol:

\Delta _{fus}H=205\frac{J}{gCu}*\frac{63.546 gCu}{1molCu}\\\\  \Delta _{fus}H=13026.93J/mol=13.03kJ/mol

Regards!

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Remember pH=-log(H ions). So it would be pH=-log(10^-7).
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URGENT.
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Answer:

1. 72.9 atm

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Explanation:

1. Gray- lussacs law is p1/t1=p2/t2 so we use this formula to figure it out by filling in the variables and solving

p1=45.0 atm

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t2=523 K

Now we fill in this in the formula and solve - 45.0 atm/ 323 K = p2/ 523 K

and now we solve for p2 by multiplying 535k by each side to give us p2

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How many moles of Hions are present in the following aqueous solution?
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Answer:

Nitric acid is a strong acid meaning it completely dissociates in water. Therefore, you can say that the concentration of H+ ions in solution is the same as the concentration of the acid (molar concentration). So once you have that, you can use the pH formula to find the pH of the acid.

pH = -log[H+]

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Aluminum–lithium (Al–Li) alloys have been developed by the aircraft industry to reduce the weight and improve the performance of
gtnhenbr [62]

Answer:

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Explanation:

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2,42g/cm³ = 2,71g/cm³X + 0,534g/cm³Y <em>(1)</em>

<em>Where X is molar fraction of Al and Y is molar fraction of Li.</em>

X + Y = 1 <em>(2)</em>

Replacing (2) in (1):

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The concentration of Li (in wt%) is:

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3 years ago
Consider the dissolution of AB(s): AB(s)⇌A+(aq)+B−(aq) Le Châtelier's principle tells us that an increase in either [A+] or [B−]
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Answer:

A. 0.000128 M is the solubility of M(OH)2 in pure water.

B. 3.23\times 10^{-6} M is the solubility of M(OH)_2 in a 0.202 M solution of M(NO_3)_2.

Explanation:

A

Solubility product of generic metal hydroxide = K_{sp}=8.45\times 10^{-12}

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                      S         2S

The expression of a solubility product is given by :

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Solving for S:

S=0.000128 M

0.000128 M is the solubility of M(OH)2 in pure water

B

Concentration of M(NO_3)_2 = 0.202 M

Solubility product of generic metal hydroxide = K_{sp}=8.45\times 10^{-12}

M(OH)_2\rightleftharpoons M^{2+}+2OH^-

                   S          2S

So, [M^{2+}]=0.202 M+S

The expression of a solubility product is given by :

K_{sp}=[M^{2+}][OH^-]^2

8.45\times 10^{-12}=(0.202 M+S)(2S)^2

Solving for S:

S=3.23\times 10^{-6} M

3.23\times 10^{-6} M is the solubility of M(OH)_2 in a 0.202 M solution of M(NO_3)_2.

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