Ask question

Ask question

All categories

4 years ago

7

Water filter project: using different types of filters to filter water, trying to figure out if the changes to the water would b

e physical, biological or chemical.
the filters are:
rocks
charcoal
radiological
polyfill
cotton cloth.

1 answer:

Anarel [89]4 years ago

6 0

Answer:

physical

Explanation:

physical water filtration consists the use of sand, screens or cross flow filtrations.

You might be interested in

Caffeine, a stimulant in coffee and tea, has a molar mass of 194.19 g/mol and a mass percentage composition of 49.48% C, 5.19% H

lozanna [386]

Answer : The molecular formula of a caffeine is, $C_8H_{10}N_4O_2$

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{49.48g}{12g/mole}=4.12moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.19g}{1g/mole}=5.19moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{28.85g}{14g/mole}=2.06moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{16.48g}{16g/mole}=1.03moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{4.12}{1.03}=4$

For H = $\frac{5.19}{1.03}=5.03\approx 5$

For N = $\frac{2.06}{1.03}=2$

For O = $\frac{1.03}{1.03}=1$

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_4H_5N_2O_1=C_4H_5N_2O$

The empirical formula weight = 4(12) + 5(1) + 2(14) + 16 = 97 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{194.19}{97}=2$

Molecular formula = $(C_4H_5N_2O)_n=(C_4H_5N_2O)_2=C_8H_{10}N_4O_2$

Therefore, the molecular of the caffeine is, $C_8H_{10}N_4O_2$

5 0

3 years ago

Please help due today

KiRa [710]

Answer:

neodymium-145

7 0

3 years ago

Select the correct answer. Given: 2LiBr Ba → BaBr2 2Li In this chemical reaction, 325 grams of barium (Ba) react completely. How

leva [86]

The answer is b 2.37

7 0

2 years ago

How can the rate constant be determined from the rate law?

denis23 [38]

Answer:

You can determine x and y in the rate law by looking at experimental data and noticing how the change in concentration of a reactant is related to the reaction rate.

Tansley-Moderator

6 0

3 years ago

10g of a non-volatile and non-dissociating solute is dissolved in 200g of benzene.

ehidna [41]

<u>Answer: </u>The molar mass of solute is 115 g/mol.

<u>Explanation:</u>

Elevation in the boiling point is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.

The expression for the calculation of elevation in boiling point is:

$\text{Boiling point of solution}-\text{boiling point of pure solvent}=i\times K_b\times m$

OR

$\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Boiling point of pure solvent (benzene) = $80.10^oC$

Boiling point of solution = $81.20^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_b$ = Boiling point elevation constant = $2.53^oC/m$

$m_{solute}$ = Given mass of solute = 10 g

$M_{solute}$ = Molar mass of solute = ? g/mol

$w_{solvent}$ = Mass of solvent = 200 g

Putting values in equation 1, we get:

$81.20-80.10=1\times 2.53\times \frac{10\times 1000}{M_{solute}\times 200}\\\\M_{solute}=\frac{1\times 2.53\times 10\times 1000}{1.1\times 200}\\\\M_{solute}=115g/mol$

Hence, the molar mass of solute is 115 g/mol.

4 0

3 years ago