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bezimeni [28]
4 years ago
8

The density of water is 1.00 g/cm3. The density of ice is 0.92 g/cm3. By what percent is the volume increased when water is froz

en at 0°C? What is the final volume if 500 mL of water is completely frozen?
Chemistry
2 answers:
cestrela7 [59]4 years ago
8 0

Answer:

At the same temperature of 0°C, the density (mass per volume) of ice is 0.9187 gram per cubic centimeter (g cm-3 or g/cm3) while that of liquid water is 0.9998 g cm-3 (Cohen et al. 2003). The lesser density means that ice contains lesser mass (quantity of matter) per unit of volume. It also means that the molecules of water are less compressed per unit volume of ice. Another way of saying it is that ice is less compact than liquid water having the same volume.

4vir4ik [10]4 years ago
7 0

Answer:

Volume increase by 8.70% and volume of 500 mL of water when completely frozen is 543 mL

Explanation:

  • Density is the ratio of mass to volume
  • Let's say there are m g of water. So volume of m g of water is equal to \frac{m}{1.00}cm^{3} or m cm^{3}
  • If m g of water is completely frozen to ice then volume of m g of ice is equal to \frac{m}{0.92}cm^{3}
  • So, percentage of increases in volume = \frac{(\frac{m}{0.92})-(m)}{m}\times 100 % = 8.70 %
  • Mass of 500 mL of water = (1.00g/cm^{3})\times 500cm^{3}=500g (1mL=1cm^{3})
  • So volume of 500 mL of water when it is completely frozen = \frac{500g}{0.92g/cm^{3}}=543cm^{3}=543 mL
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The true absorbance for a 1.0 x 10 −5 M solution is 0.7526. If the percentage stray light for a spectrophotometer is 0.56%, calc
Korvikt [17]

Answer:

The percentage deviation is  \Delta M = 1.87%

Explanation:

From the question we are told that  

     The concentration is of the solution is C = 1.0*10^{-5} M

     The true absorbance A = 0.7526

      The percentage of transmittance due to stray light z = 0.56% =\frac{0.56}{100}  = 0.0056

Generally Absorbance is mathematically represented as

           A = -log T

Where T is  the percentage of true transmittance

    Substituting value  

          0.7526 = - log T

              T = 10^{-0.7526}

                  = 0.177

                  = 17.7%

The Apparent absorbance is mathematically represented

           A_p = -log (T +z)

Substituting values

           A_p = -log(0.177 + 0.0056)

                = -log(0.1826)

               = 0.7385

The percentage by which apparent absorbance deviates from known absorbance is mathematically evaluated as

       \Delta A = \frac{A -A_p}{A} * \frac{100}{1}

              = \frac{0.7526 - 0.7385}{0.7526} * \frac{100}{1}

             \Delta A = 1.87%  

Since Absorbance varies directly with concentration the percentage deviation of the apparent concentration from know concentration  is

              \Delta M = 1.87%

           

6 0
4 years ago
Your lab partner accidentally trips and douses the front of your lab coat and shirt with a hazardous material. The TA immediatel
Alja [10]

Answer:

Remove your shirt and any other clothing that were in contact with the chemical

Explanation:

A hazardous chemical has the capacity to cause damage to the body when it comes in contact with the skin. Many hazardous chemicals are capable of being absorbed into the body via the skin.

Once your lab coat is already doused with the chemical and it has started soaking into you clothing, you must remove your lab coat, shirt and all clothing that were in contact with the hazardous chemical as a safety measure to avoid absorption of the chemical into the body via the skin. Some of these chemicals also cause damages directly to the skin and this must be avoided.

8 0
3 years ago
What is the mass in grams of 6.022 x 1023 molecules of CO2?
Kamila [148]

Answer: Molar mass of CO2 is 44 gram/mol. So,the mass of 1 mole or 6.02*10^23 molecules of CO2 is 44 grams

Explanation:

4 0
3 years ago
What volume of 1.27 M HCl is required to prepare 197.4 mL of 0.456 M HCl
Orlov [11]

Answer:

70.88 mL volume of 1.27 M of HCl is required.

Explanation:

Given data:

Initial volume = ?

Initial  molarity =  1.27 M

Final volume = 197.4 mL

Final molarity = 0.456 M

Solution:

Formula:

M₁V₁ = M₂V₂

Now we will put the values in formula.

1.27 M × V₁ =  0.456 M × 197.4 mL

V₁ = 0.456 M × 197.4 mL/1.27 M

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V₁ = 70.88 mL

70.88 mL volume of 1.27 M of HCl is required.

7 0
3 years ago
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Ivahew [28]

Answer: 2.54g

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1mole (34g) of H2O2 contains 6.02x10^23 molecules

Therefore Xg of H2O2 will contain 4.5x10^22 molecules i.e

Xg of H2O2 = (34x4.5x10^22)/6.02x10^23 = 2.54g

5 0
4 years ago
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