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Fiesta28 [93]
3 years ago
9

Show the conversion factor from cm2/stoft2/hris 3.875.

Chemistry
1 answer:
lys-0071 [83]3 years ago
6 0

Explanation:

In 1 centimeter square there are 0.00107639 square feets.

1 cm^2=0.00107639 ft^2

In 1 hour there 3600 seconds.

1 seconds = 0.000277778 Hours

1cm^2/s=(x)\times ft^2/h..(1)

1 cm^2/h=\frac{1\times 0.00107639 ft^2}{1\times 0.000277778 h}

=3.975000899 ft^2/h\approx 3.875 ft^2/h

On comparing with (1) we get:

1 cm^2/s=(3.875)\times ft^2/h

3.875 is the conversion factor.

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8. A sample of chloroform is found to contain 24.0 g of carbon,212.8 g of chlorine, and 2.02 g of hydrogen. If a second sample o
Misha Larkins [42]

Answer:

595.5

Explanation:

chloroform with 24.0 g C was 238.2 g

24g/238.2g= 60g/x

595.5

4 0
3 years ago
What is milliliter the same as?
Klio2033 [76]

Answer:

1 cubic centimeter

Explanation:

8 0
3 years ago
Read 2 more answers
For lunch, a patient consumed 3 oz of skinless chicken, 3 oz of broccoli, 1 medium apple, and 1 cup of nonfat milk (see Table 3.
Volgvan

Lunch of a patient has 3 oz skinless chicken, 3 oz of broccoli, 1 medium apple, and 1 cup of nonfat milk

Energy content of 3 oz skinless chicken is = 110 kcal

Energy content of 3 oz broccoli = 30 kcal

Energy content of 1 medium apple = 60 kcal

Energy content of 1 cup non-fat milk = 90 kcal

So the kilocalories of energy patient obtained from lunch

                   = 110 kcal+ 30 kcal + 60 kcal + 90 kcal = 290 kcal


3 0
3 years ago
Fill in the coefficients to balance the equation for the chemical reaction that occurs:
Dominik [7]

Answer:

The equation is balanced

Explanation:

NaCl (aq) + AgNO3(aq) ––> AgCl (s) + NaNO3 (aq)

NaCl (aq) + AgNO3 (aq)

Na = 1 , Cl=1 , Ag = 1 , No3= 1

AgCl (s) + NaNO3 (aq)

Ag = 1 , Cl=1 , Na = 1 , No3= 1

8 0
2 years ago
Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25
Veseljchak [2.6K]

Answer: The molecular formula will be C_6H_6O_6

Explanation:

Mass of CO_2 = 17.95 g

Mass of H_2O= 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, =\frac{12}{44}\times 17.95=4.89g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, =\frac{2}{18}\times 4.87=0.541g of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H =  0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.89g}{12g/mole}=0.407moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.541g}{1g/mole}=0.541moles

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{6.57g}{16g/mole}=0.410moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.407}{0.407}=1

For H =\frac{0.541}{0.407}=1

For O = \frac{0.410}{0.407}=1

The ratio of C : H : O = 1: 1  : 1

Hence the empirical formula is CHO.

Hence the empirical formula is CHO

The empirical weight of CHO = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of = \frac{44.0}{0.25}\times 1=176g

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{176g}{29g}=6

The molecular formula will be=6\times CHO=C_6H_6O_6

5 0
3 years ago
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