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3 years ago

The diagram shows the direction of oxygen transfer from red blood cells to body cells.

Chemistry

1 answer:

jolli1 [7]3 years ago

4 0

Answer:

The diagr displays the "DIFFUSION" of oxygen from the red blood cells into the capillary to the body cells

You might be interested in

What does the symbol H stand for

muminat

Answer:

enthalpy

Explanation:

If it is used with a triangle in front, (delta H), that means the change in enthalpy. Delta H= (m)(s)(Delta T). m=mass of products, s=heat of the products, Delta T = change in temperature.

Hope that helps

4 0

3 years ago

7.7. A painter leans his back against a painted wall while looking into a 1m long mirror at the opposite end of a rectangular ro

Karolina [17]

Answer:

Correct answer is b) 2m

Explanation:

8 0

3 years ago

2C_H. + 702 — 400, + 6H2O

astra-53 [7]

Balanced Eqn

By the Balanced eqn

60g ethane requires 7x32= 224g oxygen

here ethane is in excess.oxygen will be fully consumed

hence

300g oxygen will consume

⋅

300

224

80.36

ethane

leaving (270-80.36)= 189.64 g ethane.

By the Balanced eqn

60g ethane produces 4x44 g CO2

hence amount of CO2 produced =

⋅

80.36

235.72

and its no. of moles will be

235.72

=5.36 where 44 is the molar mass of Carbon dioxide

hope this helps

6 0

3 years ago

18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c

Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

Percentage by mass of hydrogen: 0.77%

b) Percentage by mass of chlorine: 80.37%

c) Molecular formula: $C_{2} H Cl_{3}$

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

$0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}$

The same process is used to calculate the amount of hydrogen:

$0.089g H_{2}O*\frac{2g H}{18g H_{2}O } = 0.010g H$

The percentage by mass of carbon and hydrogen are calculated as follows:

%C $\frac{0.238g}{1.3g} *100%$ = 18.3%

%H $\frac{0.010g}{1.3g} *100%$ =0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

$1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}$ = 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl= $\frac{0.43g}{0.535g} * 100%$ = 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

$18.3g C*\frac{1mol C}{12g C} = 1.52mol C$

$0.77g H*\frac{1mol H}{1g H} = 0.77mol H$

$80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl$

Dividing each of the quantities above by the smallest (0.77mol), the subscripts in a tentative formula would be

$C=\frac{1.52}{0.77} = 1.97$ ≈ 2

$H = \frac{0.77}{0.77} = 1$

$Cl =\frac{2.27}{0.77}=2.94$ ≈3

The empirical formula for the compound is:

$C_{2} H Cl_{3}$

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0

3 years ago

B) How many kilojoules of heat will be released by the combustion of 22.52 g of this liquid at

snow_tiger [21]

Answer:

Explanation:

You realize that C2H5OH releases -1277.3kJ/mol. We need to convert this to the amount based on the question. We that 22.52g of C2H5OH = 0.48884 mol.

This means that it will release (-1277.3)(0.48884) = 624.40 KJ of heat will be released. Note the negative sign is not necessary here (I think) because it says how much is released and not the change in heat of the system so it should be positive.

3 0

2 years ago