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3 years ago

8

You are planting flowers in large pots. You plan to plant flowers in 10 pots. 15 cups of potting soil will fill one pot. A bag o

f potting soil contains 25 cups. How many bags of potting soil do you need? show work

1 answer:

dimaraw [331]3 years ago

3 0

Answer:

I will need six (6) bags of potting soil

Explanation:

Since you plan on planting in 10 pots and need 15 cups of potting soil per pot, the total amount of potting soil you need <em>(in cups)</em> is 10 X 15 = 150 cups of potting soil.

We have that a bag contains 25 sups. to get the number of bags needed, we have to divide 150 by 25. This will give us 150 / 25 = 6 bags.

Therefore, I will need six (6) bags of potting soil

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My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h

Fed [463]

Answer:

P=atm

$b=\frac{L}{mol}$

Explanation:

The problem give you the Van Der Waals equation:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

First we are going to solve for P:

$(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}$

$P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}$

Then you should know all the units of each term of the equation, that is:

$P=atm$

$n=mol$

$R=\frac{L.atm}{mol.K}$

$a=atm\frac{L^{2}}{mol^{2}}$

$b=\frac{L}{mol}$

$T=K$

$V=L$

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

$P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}$

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

$P=\frac{L.atm}{L-L}-atm$

Then operate the fraction subtraction:

P= $P=\frac{L.atm-L.atm}{L}$

$P=\frac{L.atm}{L}$

And finally you can find the answer:

P=atm

Now solving for b:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

$(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}$

Replacing units:

$b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}$

Multiplying and dividing units,(please see the second photo below), we have:

$b=\frac{L-\frac{L.atm}{atm}}{mol}$

$b=\frac{L-L}{mol}$

$b=\frac{L}{mol}$

7 0

2 years ago

4 HF(g)+SiO2(s) → SiF4(9)+2 H2O(9) <br> Is the Si oxidized or reduced?

Airida [17]

Answer:

Si is reduced since it loses the oxygen atom

8 0

3 years ago

A student was asked to prepare 500.0 mL of 6.0 M NaOH. The student measured 120.0 g of NaOH and placed it in a 1000 mL beaker. T

Morgarella [4.7K]

Answer:

The concentration of the solution will be much lower than 6M

Explanation:

To prepare a solution of a solid, the appropriate mass is taken and accurately weighed in a weighing balance and then made up to mark with distilled water.

From

n= CV

n = number of moles m/M( m= mass of solid, M= molar mass of compound)

C= concentration of substance

V= volume of solution

m=120g

M= 40gmol-1

V=500ml

120/40= C×500/1000

C= 120/40× 1000/500

C=6M

This solution will not be exactly 6M if the student follows the procedure outlined in the question. The actual concentration will be much less than 6M.

This is because, solutions are prepared in a standard volumetric flask. Using a 1000ml beaker, the student must have added more water than the required 500ml thereby making the actual concentration of the solution less than the expected 6M.

7 0

3 years ago

there are tiny plants growing on rock fence after the several year . you observe something happened on the rock which of the fol

Bingel [31]

it is moss

Explanation:

5 0

2 years ago

All the substances listed below are fertilizers that contribute nitrogen to the soil.

kvv77 [185]

Answer:

Ammonia is the richest source of nitrogen on a mass percentage basis because it has 82.35% of nitrogen by mass.

Explanation:

Percentage of element in compound :

$=\frac{\text{number of atoms}\times text{Atomic mass}}{\text{molar mas of compound}}\times 100$

(a) Urea, $(NH_2)_2CO$

Molar mass of urea = 60 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 2

$N\%=\frac{2\times 14 g/mol}{60 g/mol}\times 100=46.67\%$

(b) Ammonium nitrate, $NH_4NO_3$

Molar mass of ammonium nitrate = 80 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 2

$N\%=\frac{2\times 14 g/mol}{80 g/mol}\times 100=35.00\%$

(c) Nitric oxide, NO

Molar mass of nitric oxide = 30 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 1

$N\%=\frac{1\times 14 g/mol}{30 g/mol}\times 100=46.67\%$

(d) Ammonia, $NH_3$

Molar mass of ammona = 17 g/mol

Atomic mass of nitrogen = 14 g/mol

Number of nitrogen atoms = 1

$N\%=\frac{1\times 14 g/mol}{17 g/mol}\times 100=82.35\%$

Ammonia is the richest source of nitrogen on a mass percentage basis because it has 82.35% of nitrogen by mass.

4 0

3 years ago