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katrin2010 [14]

2 years ago

7

A 0.200 g sample of unknown metal x is dropped into hydrochloride acid and realeases 80.3 mL of hydrogen gas at STP using ideal

Gas law the number of miles of the unknown is

1 answer:

s344n2d4d5 [400]2 years ago

5 0

Answer:

The number of mole of the unknown metal is 3.58×10¯³ mole

Explanation:

We'll begin by calculating the number of mole hydrogen gas, H2 that will occupy 80.3 mL at stp.

This is illustrated below:

Recall:

1 mole of any occupy 22.4L or 22400 mL at stp.

1 mole of H2 occupies 22400 mL at stp.

Therefore, Xmol of H2 will occupy 80.3 mL at stp i.e

Xmol of H2 = 80.3/22400

Xmol of H2 = 3.58×10¯³ mole

Therefore, 3.58×10¯³ mole of Hydrogen gas was released.

Now, we can determine the mole of the unknown metal as follow:

The balanced equation for the reaction is given below:

X + 2HCl —> XCl2 + H2

From the balanced equation above,

1 mole of the unknown metal reacted to produce 1 mole of H2.

Therefore, 3.58×10¯³ mole of the unknown metal will also react to produce 3.58×10¯³ mole of H2.

Therefore, the number of mole of the unknown compound is 3.58×10¯³ mole.

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Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir

guajiro [1.7K]

Answer:

$y_{O2} =4.3$ %

Explanation:

The ethanol combustion reaction is:

$C_{2}H_{5} OH+3O_{2}$ → $2CO_{2}+3H_{2}O$

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

$x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}$

Dividing the previous equation by x:

$1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}$

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

$3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )$

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

$3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles$

Calculate the number of moles of CO2 and water considering the same:

$0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)$

$0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)$

The total number of moles at the reactor output would be:

$N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)$

So, the oxygen mole fraction would be:

$y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3$ %

6 0

3 years ago

Given a sample of poly[ethylene-stat-(vinyl acetate)] A.Calculate the mean repeat unit molar mass for a sample of poly[ethylene-

AlladinOne [14]

Answer:

a) The mean repeat unit molar mass for PEVA is 30.72 g/mol

b) degree of polymerization of the copolymer is 1300

Explanation:

Given that;

the wt% of copolymer consist of 12.9 wt% of vinyl acetate and 87.1 wt% Ethylene.

Basis: 100 g of PEVA consist of 12.9 of vinyl acetate and 87.1g of Ethylene.

now we calculate the mole fraction of vinyl acetate Ethylene in the copolymer;

the molecular weights of vinyl acetate and ethylene are 86.09 g/mol and 28.05 g/mol respectively

so

moles of vinyl acetate = wt. of vinyl acetate / molecular weights of vinyl acetate

moles of vinyl acetate = 12.9 g / 86.09 g/mol

moles of vinyl acetate = 0.1498 mol

moles of Ethylene = wt. of Ethylene / molecular weights of Ethylene

moles of Ethylene = 87.1 g / 28.05 d/mol

moles of Ethylene = 3.1052 mol

Total moles = 0.1498 mol + 3.1052 mol = 3.255 mol

Next we calculate the mole percent;

mole percent of vinyl acetate $X_{V}$ = moles of vinyl acetate / total moles

$X_{V}$ = (0.1498 mol / 3.255 mol) × 100

$X_{V}$ = 4.6%

mole percent of Ethylene $X_{E}$ = moles of Ethylene / total moles

$X_{E}$ = (3.1052 mol / 3.255 mol) × 100

$X_{E}$ = 95.397% ≈ 95.4%

we know that, mean repeat unit molar mass for a sample = ∑ $X_{i}$ $M_{i}$

where $X_{i}$ is the fraction ratio and $M_{i}$ is the molecular weight

so or the PEVA

mean repeat unit molar mass M = ( $X_{V}$ $M_{V}$ ) + ( $X_{E}$ $M_{E}$ )

so we substitute

M = ( 4.6% × 86.09) + ( 95.4% × 28.05 )

M = 3.96014 + 26.7597

M = 30.72 g/mol

Therefore, The mean repeat unit molar mass for PEVA is 30.72 g/mol

b)

Degree of polymerization

$DP_{n}$ = $\frac{M_{n} }{M}$

where $M_{n}$ is the number average molecular weight ( 39,870 g/mol )

so we substitute

$DP_{n}$ = 39,870 g/mol / 30.72 g/mol

$DP_{n}$ = 1297.85 ≈ 1300 { 3 significance figure }

Therefore, degree of polymerization of the copolymer is 1300

5 0

2 years ago

An unsaturated aqueous solution of NH3 is at 90.°C in 100. grams of water. According to Reference Table G, how many grams of NH3

Illusion [34]

The answer will be 2. 10.g
Hope it's help you

3 0

3 years ago

If oxygen with an electronegativity of 3.5 were to make a chemical bond with lithium, electronegativity of 1.0, what kind of bon

sergejj [24]

Answer:

Ionic.

Explanation:

Elements with higher electronegativity values are better at attracting electrons in a chemical bond.

A chemical bond is considered "ionic" if the electronegativity difference between the two bonding atoms is greater than $1.8$ .
Otherwise, this chemical bond is considered "covalent".

In this example, the difference between the electronegativity of oxygen and lithium is $3.5 - 1.0 = 2.5$ . Since $2.5 > 1.8$ , the bond between the two elements would likely be ionic.

It is possible to reach the same conclusion based on the fact that lithium is a metal while oxygen is a nonmetal. When metal elements react with non-metal elements, the product is typically an ionic compound- with ionic bonds between the atoms.

6 0

2 years ago

Consider two aqueous nonvolatile and nonelectrolyte solutions, each with a solute concentration of 1 M. One contains glucose, wh

Svetach [21]

Answer:

a and b are correct

Explanation:

This because both are aqueous solutions,therefore, identity of solvent is same that is water.

And because both solutions are non electrolyte they would not ionize in solution, and for the same concentration, the freezing point of both solution would also be same. Since depression in freezing point is a colligative property this means that it depends on number of solute particles not nature of particles .

Hence answer is that their freezing points and Identity of the solvent shall remain the same.

8 0

3 years ago