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Aleks04 [339]
3 years ago
10

A tissue sample at 275 K is submerged in 2 kg of liquid nitrogen at 70 K for cryopreservation. The final temperature of the nitr

ogen is 75 K. What is the heat capacity of the sample in J/K
Chemistry
1 answer:
Nookie1986 [14]3 years ago
3 0

Answer:

heat capacity of the sample = 37.8 J/K

Explanation:

Step 1: Data given

Temperature of the sample = 275 K

The mass of liquid nitrogen = 2kg

temperature of liquid nitrogen = 70 K

The final temperature of the nitrogen is 75 K

Step 2: Calculate heat

Q = m*c*ΔT

⇒with m = the mass of liquid nitrogen = 2 kg = 2000 grams

⇒with c= the specific heat of the liquid nitrogen = 1.04 J/g*K

⇒with ΔT = the change of temperature of liquid nitrogen = T2 - T1 = 75 - 70 = 5K

Q = 2000 grams * 1.04 J/g*K * 5K

Q = 10400 J

Step 3: Calculate the heat capacity of the sample

heat capacity of the sample = 10400 J / 275 K

heat capacity of the sample = 37.8 J/K

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When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
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Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

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A student can predict Amount of calories present if she finds the mass of whole candy bar,including the wrapper.

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Based on the mass of a whole candy bar a student can predict the calories contained in the candy bar. If the mass of candy bar is more than it means it has more calories and if the candy bar has less mass than it shows the presence of fewer calories.

More calories mean more sugar content in the candy bar. With sugar, the candy bar has carbohydrates also. The sugar present in the bar gives us energy.

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