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2 years ago

Can someone help me with the question in the image

Chemistry

1 answer:

timama [110]2 years ago

7 0

The good ozone protects us from the UV/ harmful radiations whereas bad ozone is an air pollutant.

Explanation:

There are two types of ozone layer found in the earth's atmosphere extending from troposphere to stratosphere. They are good ozone and bad ozone.
Bad ozone as mentioned earlier it is an air pollutant and found in the ground level of earth, most accurately the troposphere. Bad ozone is formed in the ground level of earth's atmosphere by the reaction between nitrogen oxides and organic compounds which are volatile
The Good ozone is found in the stratosphere layer of the earth's atmosphere. They protect us from harmful radiations. Good ozone layer in the stratosphere of the atmosphere is being destroyed by hydrocarbons, CFCs, and human intervention

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What are twoparts of an atom

Leto [7]

The answer is neutrons and protons.

Hope you got it!!

6 0

2 years ago

What is the molarity of a solution containing 56 grams of solute in 959 ml of solution? (molar mass of solute = 26 g/mol)?

jolli1 [7]

26g --- 1 mol
56g --- X
X= 56/26 = 2,154 mol

959 ml = 959cm³ = 0,959dm³

C = n/V
C = 2,154/0,959
C = 2,246 mol/dm³

8 0

2 years ago

Read 2 more answers

A compound is 54.53% c, 9.15% h, and 36.32% o by mass. what is its empirical formula? the molecular mass of the compound is 132

Yakvenalex [24]

<span>First - you need the empirical formula.

So, assume you have 100 g of the compound.

If so, you'll have 54.53 gram of C, 9.15 g of H and 36.32 g of O. Find the number of moles of each.

54.53 g C (1 mole C / 12.01 g C) = 4.540

9.15 g H (1 mole H / 1.008 g H) = 9.077

36.32 g O (1 mole O / 15.9994 g O) = 2.270

Take the smallest number found and divide the others by it to get the empirical formula.

4.540/2.270 = 2.
9.077/2.270 = 4.
2.270/2.270 =1.

So, that gives you the empirical formula of C2H4O.

Find the weight of this compound. C = 12, H = 1, O = 16. So, C2H4O is 44 amu.

132/44 = 3.

So, 3 (C2 H4 O) = C6H12O3 = molecular formula.</span>

6 0

2 years ago

An air stream enters a variable area channel at a velocity of 30 m/s with a pressure of 120 kPa and a temperature of 10°C. At a

nekit [7.7K]

Answer:

P₂= 74 kPa under constant density and ρ₂ = 1.06 kg/m³ (-38.6% of error compared with incompresible assumption) . Thus Bernoulli’s equation should not be applied

Explanation:

Assuming ideal gas behaviour of air , then

P*V= n*R*T = m / M * R *T

since

ρ= m/V = P*M /( R *T)

where

n= moles , V= volume , m= mass

ρ= density

P= pressure = 120 kPa= 120000 Pa

M= molecular weight of air = 0.21*32+0.79*28= 28.24 gr/mol = 0.02824 kg/mol

T= absolute temperature = 10°C + 273 = 283 K

R= ideal gas constant = 8.314 J/mol K

solving for ρ

ρ= P*M /( R *T) = 120000 Pa*0.02824 kg/mol/(8.314 J/mol K*283 K) = 1.47 kg/m³

then from Bernoulli's equation

P₁ + ρ*v₁²/2 = P₂ + ρ*v₂²/2

where 1 denotes inlet and 2 denotes other point , p = pressure and v= velocity . Then solving for p₂

P₁ + ρ*v₁²/2 = P₂ +ρ*P₂²/2

P₂= P₁ +ρ*v₁²/2 - ρ*v₂²/2 = P₁ +ρ/2*(v₁² - v₂²)

replacing values

P₂= P₁ +ρ/2*(v₁² - v₂²) = 120000 Pa + 1.47 kg/m³/2*[(30 m/s)²-(250 m/s)²] = 74724 Pa = 74 kPa

P₂= 74 kPa

then if the temperature remains constant

ρ₁= P₁*M /( R *T) and ρ₂= P₂*M /( R *T)

dividing both equations

ρ₂/ρ₁ = P₂/ P₁

ρ₂ = (P₂/ P₁)*ρ₁

then from Bernoulli's equation

P₁ + ρ₁*v₁²/2 = P₂ + ρ₂*v₂²/2

P₂ = P₁ + ρ₁*v₁²/2 - ρ₂*v₂²/2

therefore

ρ₂ = (P₂/ P₁)*ρ₁ = (P₁ + ρ₁*v₁²/2 - ρ₂*v₂²/2 ) /P₁ *ρ₁

P₁ * ρ₂ = P₁ *ρ₁ + ρ₁²*v₁²/2 - ρ₂*ρ₁ * v₂²/2

P₁ * ρ₂ + ρ₂*ρ₁ * v₂²/2 = P₁ *ρ₁ + ρ₁²*v₁²/2

ρ₂* (P₁ + ρ₁ * v₂²/2) = P₁ *ρ₁ + ρ₁²*v₁²/2

ρ₂ = (P₁ *ρ₁ + ρ₁²*v₁²/2)/(P₁ + ρ₁ * v₂²/2) = (P₁ + ρ₁*v₁²/2)/(P₁/ρ₁ + v₂²/2)

replacing values

ρ₂ = ( 120000 Pa + 1.47 kg/m³/2*(30 m/s)²)/(120000 Pa/1.47 kg/m³+1/2*(250 m/s)²)

ρ₂ = 1.06 kg/m³

the error of assuming constant ρ would be

e = (ρ₂ - ρ)/ρ₂= 1- ρ/ρ₂= 1- 1.47 kg/m³/1.06 kg/m³ = -0.386 (-38.6%)

thus Bernoulli’s equation should not be applied

8 0

2 years ago

A compound contains only carbon, hydrogen, and oxygen. combustion of 11.75 mg of the compound yields 17.61 mg co2 and 4.81 mg h2

Ymorist [56]

Number of moles is defined as the ratio of given mass in g to the molar mass.

First, convert the given mass of carbon dioxide in mg to g:

1 mg = 0.001 g

17.61 mg = 0.01761 g

Number of moles of carbon dioxide = $\frac{0.01761 g}{44.01 g/mol}$

= $0.0004001 mol$

Mass of carbon = number of moles of carbon dioxide \times molar mass of carbon

= $0.0004001 mol\times 12.011 g/mol$

= $0.004806 g$

Number of moles of water= $\frac{0.00481 g}{18 g/mol}$

= $2.672\times 10^{-4}$

Since, water contains two hydrogen atoms. Thus,

Moles of hydrogen = $2\times 2.672\times 10^{-4}$

= $5.34\times 10^{-4}$

Mass of hydrogen = $5.34\times 10^{-4}\times \times 1.008 g/mol$

= $5.34\times 10^{-4} g$

Mass of oxygen = $0.001175-(5.38\times 10^{-4}g+0.004806 g)$

= $0.006405 g$

Number of moles of oxygen = $\frac{0.006405 g}{15.999 g/mol}$

= $0.000400$

Now,

$C_{0.0004001} H_{0.000534} O_{0.000400}$

Divide the smallest number to get the whole number,

$C_{\frac{0.0004001}{0.000400}} H_{\frac{0.000534}{0.000400}} O_{\frac{0.000400}{0.000400}}$

we get,

$C_{1} H_{1.33} O_{1}$

Now, multiply all the subscript by 3 to get the whole number,

$C_{3} H_{4} O_{3}$ (empirical fomula)

Molar mass of the compound = $3\times 12.011 g/mol+4\times 1.008 g/mol+3\times 15.999 g/mol$

= $88.062 g/mol$

Divide given molar mass of the compound with the molar mass of the compound.

= $\frac{176.1 g/mol}{88.062 g/mol}$

= $1.999\simeq 2$

Thus, multiply the subscripts of empirical formula by 2 to get the molecular formula, we get:

$C_{6}H_{8}O_{6}$

Hence, empirical formula is $C_{3}H_{4}O_{3}$ and molecular formula is $C_{6}H_{8}O_{6}$

8 0

2 years ago