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3 years ago

Please help with this!!

Chemistry

1 answer:

loris [4]3 years ago

8 0

Answer:

Limiting: Lab covers. Books produced: 75. Excess amounts: In explanation

Explanation:

Amount that can be produced with just lab covers: 75 books (150/2)

Amount that can be produced with just lined paper: 150 books (7500/50)

Amount that can be produced with graph paper: 120 (3000/25)

Amount that can be produced with staples: Approximately 83 (250/3)

As we can see, the lab covers are limiting as they can only produce 75 books. So, we can only make 75 books.

You will have 3,750 lined paper left over (or 75 books)

(150-75=75 , 75*50=3,750)

You will have 1,125 graph paper left over (or 45 books)

(120-75=45 , 75*25=1,875 , 3000-1875=1125)

And then you will have approximately 25 staples left over (or 8 books)

(83-75=8 , 8*3=225, 250-225=25)

(Hopefully this is correct, I apologize if I messed up)

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Calculate (a) the number of moles of hydrogen required to react with 0.0969 moles of nitrogen, and (b) the number of moles of am

Bumek [7]

Answer:

The answer to your question is:

Explanation:

Data

moles H=?

moles of N = 0.0969

moles of NH₃=?

N₂ (g) + 3 H₂ (g) ⇒ 2NH₃ (g)

Process

1.- Set a rule of three to calculate the moles of hydrogen

1 mol of nitrogen ------------- 3 moles of hydrogen

0.0969 moles of N ---------- x

x = (0.0969 x 3) / 1

x = 0.2907 moles of hydrogen

2.- Set a rule of three to calculate the moles of ammonia

1 mol of nitrogen -------------- 2 moles of ammonia

0.0969 mol of N -------------- x

x = (0.0969 x 2) / 1

x = 0.1938 moles of ammonia

5 0

3 years ago

When a chemical reaction is run in aqueous solution inside a calorimeter, the temperature change of the water (and Ccal) can be

Yakvenalex [24]

Answer:

The total change in enthalpy for the reaction is - 81533.6 J/mol

Explanation:

Given the data in the question;

Reaction;

HCl + NaOH → NaCl + H₂O

Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J

Moles of NaOH = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

so, 0.0500 moles of H₂O produced

Volume of solution = 50.mL + 50.mL = 100.0 mL

Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g

now ,

Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal

we know that; The specific heat of water(H₂O) is 4.18 J/g°C.

so we substitute

q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28

q_soln = 2842.4 + 1234.28

q_soln = 4076.68 J

Enthalpy change for the neutralization is ΔH $_{neutralization}$

ΔH $_{neutralization}$ = -q_soln / mole of water produced

so we substitute

ΔH $_{neutralization}$ = -( 4076.68 J ) / 0.0500 mol

ΔH $_{neutralization}$ = - 81533.6 J/mol

Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol

6 0

3 years ago

Which process involves an increase in entropy? crystallization of a solute from a solution ice melting into liquid water iodine

GrogVix [38]

Ice melting into liquid water.

7 0

4 years ago

Read 2 more answers

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.

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3 years ago

How many electrons does it take to fill the outer energy level of most atoms

Orlov [11]

Depends on the element it can by up to 3, 8, or maybe 16.

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3 years ago