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2 years ago

How many grams are in 2.5 pound sample

Chemistry

1 answer:

julsineya [31]2 years ago

5 0

About 2,500 grams Ans balkfdoaks;

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penicillin. an important antibiotic (antibacterial agent), was discovered accidentally by the scottish bacteriologist alexander

dmitriy555 [2]

Answer:

mass percent of carbon = 57.78 %

mass percent of hydrogen = 6.40 %

mass percent of nitrogen = 8.96 %

mass percent of oxygen = 20.49 %

mass percent of sulfur = 10.24 %

Explanation:

Given data

Molecular formula = C₁₄H₂₀N₂O₄S

molecular mass (total mass) = 312.39 g/mol

Percentage of carbon = ?

Percentage of hydrogen = ?

Percentage of oxygen = ?

Percentage of nitrogen = ?

Percentage of sulfur = ?

Solution

1st we find out number of moles of each element from the molecular formula

Number of moles of carbon = 14 mol

Number of moles of hydrogen = 20 mol

Number of moles of nitrogen = 2 mol

Number of moles of oxygen = 4 mol

Number of moles of sulfur = 1 mol

Now we find out the mass of each element

as we know that

mass = number of moles × molecular mass

mass of carbon = 14 mol × 12 g/mol

mass of carbon = 168 g

mass of hydrogen = 20 mol × 1 g/mol

mass of hydrogen = 20 g

mass of nitrogen = 2 mol × 14 g/mol

mass of nitrogen = 28 g

mass of oxygen = 4 mol × 16 g/mol

mass of oxygen = 64 g

mass of sulfur = 1 mol × 32 g/mol

mass of sulfur = 32 g

now we find out the mass percent of each element

mass percent = ( mass ÷ total mass ) × 100

mass percent of carbon = ( 168 g ÷ 312.39 g/mol ) × 100

mass percent of carbon = 57.78 %

mass percent of hydrogen = ( 20 g ÷ 312.39 g/mol ) × 100

mass percent of hydrogen = 6.40 %

mass percent of nitrogen = ( 28 g ÷ 312.39 g/mol ) × 100

mass percent of nitrogen = 8.96 %

mass of oxygen =( 64 g ÷ 312.39 g/mol ) × 100

mass percent of oxygen = 20.49 %

mass percent of sulfur = ( 32 g ÷ 312.39 g/mol ) × 100

mass percent of sulfur = 10.24 %

7 0

2 years ago

How many grams of Boron can be obtained from 234 grams of B2O3?

Xelga [282]

Answer:

72.67g of B

Explanation:

The reaction of B₂O₃ to produce boron (B), is:

B₂O₃ → 3/2O₂ + 2B

That means B₂O₃ produce 2 moles of boron

Molar mass of B₂O₃ is 69.62g/mol. 234g of B₂O₃ contains:

234g B₂O₃ ₓ (1mol / 69.62g) = 3.361 moles of B₂O₃.

As 1 mole of B₂O₃ produce 2 moles of B, Moles of B that can be produced from B₂O₃ is:

3.361mol B₂O₃ ₓ 2 = 6.722 moles of B.

As molar mass of B is 10.811g/mol. Thus mass of B that can be produced is:

6.722mol B ₓ (10.811g / mol) = 72.67g of B

4 0

2 years ago

At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5

Annette [7]

Answer:

4600s

Explanation:

$2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}$

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:

$-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt$

If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.

PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.

$-\frac{d[P(B)]}{P(B)}=k*dt$

$-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt$

Integrating we get:

$\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt$

$-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})$

Clearing for t2:

$\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}$

$ln[P(N_{2}O_{5})]=ln(650)=6.4769$

$ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333$

$t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s$

4 0

2 years ago

Diborane, B2H6 a possible rocket propellant, can be made by using lithium hydride (LiH): 6 LiH+ 2 BCl2àB2H6+ 6 LiCl . If you mix

Genrish500 [490]

Answer :

(a) Limiting reactant = $LiH$

(b) The excess reactant = $BCl_3$

(c) The percent of excess reactant is, 50.87 %

(d) The percent yield of $B_2H_6$ or percent conversion of $LiH$ to $B_2H_6$ is, 38.80 %

(e) The mass of $LiCl$ produced is, 1066.42 lb

Explanation : Given,

Mass of $LiH$ = 200 lb = 90718.5 g

conversion used : (1 lb = 453.592 g)

Mass of $BCl_3$ = 1000 lb = 453592 g

Molar mass of $LiH$ = 7.95 g/mole

Molar mass of $BCl_3$ = 117.17 g/mole

Molar mass of $B_2H_6$ = 27.66 g/mole

Molar mass of $LiCl$ = 42.39 g/mole

First we have to calculate the moles of $LiH$ and $BCl_3$ .

$\text{Moles of }LiH=\frac{\text{Mass of }LiH}{\text{Molar mass of }LiH}=\frac{90718.5g}{7.95g/mole}=11411.13moles$

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{453592g}{117.17g/mole}=3871.23moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$6LiH+2BCl_3\rightarrow B_2H_6+6LiCl$

From the balanced reaction we conclude that

As, 6 moles of $LiH$ react with 1 mole of $BCl_3$

So, 11411.13 moles of $LiH$ react with $\frac{11411.13}{6}=1901.855$ moles of $BCl_3$

From this we conclude that, $BCl_3$ is an excess reagent because the given moles are greater than the required moles and $LiH$ is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 3871.23 - 1901.855 = 1969.375 moles

Total excess reactant = 3871.23 moles

Now we have to determine the percent of excess reactant $(BCl_3)$ .

$\% \text{ excess reactant}=\frac{\text{Moles of remaining excess reactant}}{\text{Moles of total excess reagent}}\times 100$

$\% \text{ excess reactant}=\frac{1969.375}{3871.23}\times 100=50.87\%$

The percent of excess reactant is, 50.87 %

Now we have to calculate the moles of $B_2H_6$ .

As, 6 moles of $LiH$ react to give 1 mole of $B_2H_6$

So, 11411.13 moles of $LiH$ react to give $\frac{11411.13}{6}=1901.855$ moles of $B_2H_6$

Now we have to calculate the mass of $B_2H_6$ .

$\text{Mass of }B_2H_6=\text{Moles of }B_2H_6\times \text{Molar mass of }B_2H_6$

$\text{Mass of }B_2H_6=(1901.855mole)\times (27.66g/mole)=52605.3093g$

Now we have to calculate the percent yield of $B_2H_6$ .

$\%\text{ yield of }B_2H_6=\frac{\text{Actual yield of }B_2H_6}{\text{Theoretical yield of }B_2H_6}\times 100=\frac{20411.7g}{52605.3093g}\times 100=38.80\%$