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jonny [76]
3 years ago
8

Can someone please help me with these three questions, it will soon be due.

Chemistry
1 answer:
Mice21 [21]3 years ago
8 0

YOU also better mark me BRAINLIEST if someone else answers!!!

1)

volume = 100 mL

density = 1.02 g / mL

mass = density x volume

  • = 1.02 x 100
  • = 102 g
  1. Cp = 4.03 J / oc g
  2. dT = 28.7 - 23.5 = 5.2 oC
  3. Q = m Cp dT
  4. Q = 102 x 4.03 x 5.2
  5. Q = 2138 J
  6. Q = 2.14 kJ

heat absorbed = 2.14 kJ

2)

  • We can calculate the moles of HC1 present in 50 ml 1m solution by using the formula, moles = volume of solution  in L*molarity
  • moles = 50/1000 = .05 , L*1 M = .05 Moles of HCL.
  • The same can be applied for calculation  of moles fo  Na OH of 50 ml, 1 M solution. (50 ml NaOH = .o5 L NaOH)
  • Then use the formula = volume of solution / 1000*moles.wt,
  • =1.825 gram.
  • moles=mass in grams /moles.wt= .05 moles of HCI
  • calculate moles of water from this reaction.
  • HC1+NaOH----->NACl+h20 (Balanced Equation)
  • .5+.5----> .5+.5
  • .05 moles of water
  • Delta H=q/n.
  • n=.05 to calculate the Delat H.

3) I don't understand it, sorry


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Explanation:

THE QUESTION SHOULD BE

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Write the mathematical form of the concentration of the products, then divide it by the concentration of the reactants as follows

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4 years ago
Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is no
nasty-shy [4]

This question is incomplete, the complete question is;

Tonksite is a solid at 300.00K.  At 300.00 K its enthalpy of sublimation is 66.00 kJ/mol. The sublimation pressure at 300.00 K is 5.00 × 10⁻⁴ atm

Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is not a function of temperature.

Answer: the sublimation pressure of the solid at the melting point is  0.3727 atm

Explanation:

Given that;

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T2 = 400 K

H_sub = 66 kJ/mol = 66000 J/mol

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p2 = ?

now using the expression

log( p2 / 5.00 × 10⁻⁴ ) = (H_sub / R ×  2.303 ) (( T2 - T1) / T1T2)

now we substitute of given values into the expression

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3 years ago
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Answer:

64.52 mg.

Explanation:

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Time (t) = 1000 years.

Final amount (N) =.?

Next, we shall determine the rate constant (K).

This is illustrated below:

Half life (t½) = 1590 years

Rate/decay constant (K) =?

K = 0.693 / t½

K = 0.693/1590

K = 4.36×10¯⁴ / year.

Finally, we shall determine the amount that will remain after 1000 years as follow:

Half life (t½) = 1590 years

Initial amount (N₀) = 100 mg

Time (t) = 1000 years.

Rate constant = 4.36×10¯⁴ / year.

Final amount (N) =.?

Log (N₀/N) = kt/2.3

Log (100/N) = 4.36×10¯⁴ × 1000/2.3

Log (100/N) = 0.436/2.3

Log (100/N) = 0.1896

Take the antilog

100/N = antilog (0.1896)

100/N = 1.55

Cross multiply

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Divide both side by 1.55

N = 100/1.55

N = 64.52 mg

Therefore, the amount that remained after 1000 years is 64.52 mg

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Answer:

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Explanation:

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