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PtichkaEL [24]
3 years ago
15

If a 28.0 L balloon with a temperature of

Chemistry
1 answer:
Alexeev081 [22]3 years ago
6 0

Answer:

5.6 L

Explanation:

We can apply Charles' Law here since our pressure is constant (will not change inside the refrigerator) and we are relating change in volume with change in temperature:

V₁ / T₁ = V₂ / T₂ where V₁ and T₁ are initial volume and temperature, and V₂ and T₂ are final volume and temperature. Let's plug in what we know and solve for the unknown:

28.0 L / 25 °C = V₂ / 5 °C => V₂ = 5.6 L

5.6 L is our new volume (at 5 °C).

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Answer:

See explanation

Explanation:

Salt makes ice to melt faster than it usual. Sugar is another substance that influences the melting time of ice.

If i have a sample of pure ice and another sample of ice treated with sugar. I have to maintain the both initially at 0 degrees temperature and have a stop clock to measure the melting time.

I will ultimately notice that the sample of ice treated with sugar will melt faster than the pure sample of ice due to the presence of an impurity.

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Be sure to answer all parts. Dimercaprol (HSCH2CHSHCH2OH) was developed during World War I as an antidote to arsenic-based poiso
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<u>Answer:</u>

<u>For A:</u> The number of arsenic atoms are 3.4\times 10^{21}

<u>For B:</u> The percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

<u>Explanation:</u>

  • <u>For A:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of dimercaprol = 696 mg = 0.696 g    (Conversion factor:  1 g = 1000 mg)

Molar mass of dimercaprol = 124.21 g/mol

Putting values in above equation, we get:

\text{Moles of dimercaprol}=\frac{0.696g}{124.21g/mol}=0.0056mol

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of molecules.

So, 0.0056 moles of dimercaprol will contain 0.0056\times 6.022\times 10^{23}=3.4\times 10^{21} number of molecules.

As, 1 molecule of dimercaprol binds with 1 atom of Arsenic

So, 3.4\times 10^{21} number of dimercaprol molecules will bind with = 1\times 3.4\times 10^{21}=3.4\times 10^{21} number of arsenic atoms

Hence, the number of arsenic atoms are 3.4\times 10^{21}

  • <u>For B:</u>

We know that:

Molar mass of dimercaprol = 124.21 g/mol

Molar mass of mercury = 200.59 g/mol

Molar mass of thallium = 204.38 g/mol

Molar mass of chromium = 51.99 g/mol

Also, 1 molecule of dimercaprol binds with 1 metal atom.

To calculate the percentage composition of metal in a complex, we use the equation:

\%\text{ composition of metal}=\frac{\text{Mass of metal}}{\text{Mass of complex}}\times 100     ......(1)

  • <u>For mercury:</u>

Mass of Hg-complex = (200.59 + 124.21) = 324.8 g

Mass of mercury = 200.59 g

Putting values in equation 1, we get:

\%\text{ composition of mercury}=\frac{200.59g}{324.8g}\times 100=61.76\%

  • <u>For thallium:</u>

Mass of Tl-complex = (204.38 + 124.21) = 328.59 g

Mass of thallium = 204.38 g

Putting values in equation 1, we get:

\%\text{ composition of thallium}=\frac{204.38g}{328.59g}\times 100=62.2\%

  • <u>For chromium:</u>

Mass of Cr-complex = (51.99 + 124.21) = 176.2 g

Mass of chromium = 51.99 g

Putting values in equation 1, we get:

\%\text{ composition of chromium}=\frac{51.99g}{176.2g}\times 100=29.51\%

Hence, the percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

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Answer:

<h2>Dog's mitochondria lack the transport protein that transport  pyruvate ( end product of glycolysis) across the outer mitochondrial  membrane .</h2>

Explanation:

1. As given here that dog's mitochondria can use only fatty acids and also  amino acids for their respiration, and  as compared to others, Dong's cell produce more lactate then normal,  this indicate that his mitochondrial membrane is different then others.  

2. The aerobic phases of cellular respiration in eukaryotes occur within  mitochondria. These aerobic phases are the TCA Cycle and the electron transport chain. Glycolysis occurs in the cytoplasm and the products of glycolysis enter into the mitochondria to continue cellular respiration.

3. These condition shows that dog's mitochondria lack the transport protein of mitochondria that moves pyruvate across the outer mitochondrial  membrane.

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