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3 years ago

as the temperature on a sample of gas with a constant volume increases, the pressure _____. decreases stays the same increases

1 answer:

3 0

The pressure increases as temperature increases, as a gas is heated it expands, so in a confined container (constant volume) pressure must increase, this can be observed through the universal gas law (pv=nRT)

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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

2Fe(s) +3H2SO4(aq) →Fe2(SO4)3(aq) +3H2(g)When 10.3 g of iron are reacted with 14.8 moles of sulfuric acid, what is the percent y

Elden [556K]

Answer:

1040%

Explanation:

To solve this question we must convert the mass of Iron to moles in order to find limiting reactant. With limiting reactant we can find the theoretical moles of hydrogen and theoretical mass:

Percent yield = Actual yield (5.40g) / Theoretical yield * 100

Moles Fe -Molar mass: 55.845g/mol-:

10.3g * (1mol / 55.845g) = 0.184 moles of Fe will react.

For a complete reaction of these moles there are necessaries:

0.184 moles Fe* ( 3 mol H2SO4 / 2 mol Fe) = 0.277 moles H2SO4.

As there are 14.8 moles of the acid, Fe is limiting reasctant.

The moles of H2 produced are:

0.184 moles Fe* ( 3 mol H2 / 2 mol Fe) = 0.277 moles H2

The mass is:

0.277 moles H2 * (2.016g/mol) = 0.558g H2

Percent yield is:

5.40g / 0.558g * 100 = 1040%

It is possible the experiment wasn't performed correctly

7 0

3 years ago

A basketball weighs approximately 1.35 kg. What is this mass in grams (g)? step by step

choli [55]

Answer:

1350 g

Explanation: just add a 0

8 0

3 years ago

24-Complete the following sentence:

docker41 [41]

Answer:

B. Solvent

Explanation:

In osmosis, water always moves from low solute concentration to high solute concentration. SOLUTE NEVER MOVES AS IT CANNOT PASS THE SELECTIVELY PERMEABLE MEMBRANE. alot of caps but need to stress this concept cuz otherwise this concept gets very confusing

8 0

3 years ago

For a particular redox reaction, NO-2 is oxidized to NO-3 and Ag+ is reduced to Ag . Complete and balance the equation for this

Salsk061 [2.6K]

The following are the steps to complete and balnce the equation for the given reaction

Explanation:

We are given, NO2– is oxidized to NO3– and Ag is reduced to Ag

NO2– + Ag+ -----> NO3– + Ag(s)

Step 1) Assign the oxidation state to each element reaction

NO2– + Ag+ -----> NO3– + Ag(s)

N= +3 N = +5

O = -2 O = -2

Ag = +1 Ag = 0

NO2– -----> NO3– ………oxidation half reaction

Ag+ -----> Ag(s) ……….reduction half reaction

Step 2) Balance the element other than O and H

NO2– -----> NO3–

Ag+ -----> Ag(s)

Step 3) Balance the O by adding 1 H2O for 1 O

NO2– + H2O -----> NO3–

Ag+ -----> Ag(s)

Step 4) Balance the H by adding H+

NO2– + H2O -----> NO3– + 2H+

Ag+ -----> Ag(s)

Step 5) Balance the charge by adding electron

NO2– + H2O -----> NO3– + 2H+ + 2e-

Ag+ + 1e------> Ag(s)

Step 6) Balance the electron in both half reaction

NO2– + H2O -----> NO3– + 2H+ + 2e-

2 Ag+ + 2e------> 2 Ag(s)

3 0

3 years ago