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Tju [1.3M]
3 years ago
14

What mass of water absorbs 6700 J of heat to raise the temperature from 283K to 318K?​

Chemistry
1 answer:
larisa86 [58]3 years ago
4 0

Answer:

you did not give the specific heat like formula like it takes 1kj to raise 28grams of water by 10 grams

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In a titration, 25.0 mL of 0.500 M KHP is titrated to the equivalence point with NaOH. The final solution volume is 53.5 mL. Wha
Lena [83]
M1v1=m2v2 
m2=(m1v1)/v2 
Where m is the molarities and v is the volumes
<span>m2=(25.0*0.500)/53.5
m2=12.5/53.5
m2=0.2336
by rounding off:
m2=0.234 M
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3 years ago
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What molecules do not attract each other
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Answer:

Non-polar and non-polar molecules do not attract each other.

Explanation:

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What is the total mass of D-glucose dissolved in a 2-μL aliquot of the solution used for this experiment?
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Answer:

The total mass of D-Glucose dissolved in a 2μL aliquot is 1 E-4 g

Explanation:

providing a solution to 5% weight-volume as found in commerce:

⇒ % 5 = (5g d-glucose/ 100 mL sln)×100

⇒ 0.05 =  g C6H12O6/mL sln

⇒ g C6H12O6 = (2 μL sln)×(0.001 mL/μL)×(0.05 g C6H12O6/mL sln)

⇒ g C6H12O6 = 1 E-4 g C6H12O6

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4 years ago
N2 + 3H2 mc011-1.jpg 2NH3<br> What is the mole ratio of hydrogen to ammonia?
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1 N₂ + 3 H₂ = 2 NH₃

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Calculate Ho298 for the process
Inga [223]

Explanation:

As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

Sb + \frac{3}{2}Cl_2 \rightarrow SbCl_{3}    \Delta H^0_1 =  -314 kJ  ..........(1)

SbCl_{3} + Cl_2 \rightarrow SbCl_{5}    \Delta H^0_2 = -80kJ   ..............(2)

The final reaction is as follows:  

Sb + \frac{5}{2}Cl_{2} \rightarrow SbCl_{5}  \Delta H^0_3 = ?  .............(3)

Therefore, adding (1) and (2) we get the final equation (3) and value of \Delta H^{0}_{3} at 298 K will be as follows.

             \Delta H^{0}_{3} = \Delta H^{0}_{1} + \Delta H^{0}_{2}    

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Thus, we can conclude that H^{o} at 298 K for the given process is -394 kJ.

4 0
3 years ago
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