Ask question

Ask question

All categories

3 years ago

9

What factors affect water quality?In a typical drinking-water treatment process, ____________________ is when air is forced thro

ugh the water to reduce unpleasant odors and taste.

1 answer:

ki77a [65]3 years ago

3 0

Aeration is the answer to your question

You might be interested in

Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon

Pani-rosa [81]

<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

<u>Explanation:</u>

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: $Cu_2O\text{ and }CuO$

<u>For Sample 1:</u>

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = $\frac{2.26}{2.26}=1$

For Oxygen = $\frac{4.55}{2.26}=2.01\approx 2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is $CO_2$

<u>For Sample 2:</u>

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = $\frac{3.57}{3.57}=1$

For Oxygen = $\frac{3.57}{3.57}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is $CO$

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = $\frac{1}{2}\times 1=\frac{1}{2}$

Now, taking the ratio of carbon atoms in both the samples, we get:

$C_1:C_2=\frac{1}{2}:1=1:2$

Hence, the ratio of carbon in both the compounds is 1 : 2

8 0

3 years ago

Nadusha1986 [10]

If I am correct only 1

7 0

3 years ago

Read 2 more answers

We have two surfaces A and B where same amount of Force is acting. Area of Surface A is greater than Area of Surface B. which of

Phoenix [80]

Answer:

Pressure on Surface A > Pressure on Surface B

Explanation:

Pressu

re on Surface B > Pressure on Surface A

7 0

3 years ago

What is the energy of a photon that emits a light of frequency 6.42 x 1014 Hz?<br><br>

svetlana [45]

Answer:

Option B. 4.25×10¯¹⁹ J

Explanation:

From the question given above, the following data were obtained:

Frequency (f) = 6.42×10¹⁴ Hz

Energy (E) =?

Energy and frequency are related by the following equation:

Energy (E) = Planck's constant (h) × frequency (f)

E = hf

With the above formula, we can obtain the energy of the photon as follow:

Frequency (f) = 6.42×10¹⁴ Hz

Planck's constant (h) = 6.63×10¯³⁴ Js

Energy (E) =?

E = hf

E = 6.63×10¯³⁴ × 6.42×10¹⁴

E = 4.25×10¯¹⁹ J

Thus, the energy of the photon is 4.25×10¯¹⁹ J

3 0

3 years ago

Read 2 more answers

By what factor would the average kinetic energy of the particles by increased if the temperature of a substance was increased fr

natita [175]

Average kinetic energy of a particle :
0.5 mv^2 = kT/2
so the kinetic energy = kT/2
assuming the same value of K
T1 = -49 + 273 = 224
T2 = 287 + 273 = 560

E2 / E1 = kT2 / 2 / kT1 / 2
E2 / E1 = T2 / T1
E2 / E1 = 560 / 224 = 2.5
so the average kinetic energy of the particle increases by 2.5

3 0

3 years ago