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erma4kov [3.2K]
3 years ago
10

What poisonous gas is found in the exhaust fumes of car engines?

Chemistry
2 answers:
Xelga [282]3 years ago
6 0
The poisonous gases found in the exhaust fumes of car engines are the products of combustion which are carbon dioxide and carbon monoxide. They are known to be toxic and carcinogenic. Hope this answers the question. Have a nice day.
Ugo [173]3 years ago
4 0
The toxic gar expelled from the reaction between gasoline and oxygen in the vehicle's engine is both Carbon dioxide and monoxide
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respiratory and lymphatic

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Complete the sentence indicating the number of protons and electrons in the given ion. Match the numbers in the left column to t
Cloud [144]

Answer: (Chlorine-35)

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5 0
3 years ago
What's the formula mass for (NH4)3PO4
maria [59]
The formula mass of a molecule is the sum of the atomic weights of the atoms in the empirical formula of the compound. It is also known as Formula Weight.
The atomic weights of
N = 14.01 amu
H = 1.00 amu
P = 30.97 amu
O = 16.0 amu

Now, we will calculate now the formula mass of a given substance
3(14.01) + 12(1.00) + 1(30.97) + 4(16.0) = 42.03 + 12.00 + 30.97 + 64.0 = 149.0 amu

Therefore, the formula mass for (NH4)3PO4 is 149.0 amu

5 0
3 years ago
Given that the freezing point depression constant for water is 1.86°c kg/mol, calculate the change in freezing point for a 0.907
melamori03 [73]

Answer : The correct answer for change in freezing point = 1.69 ° C

Freezing point depression :

It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m

Where : ΔTf = change in freezing point (°C)

i = Von't Hoff factor

kf =molal freezing point depression constant of solvent.\frac{^0 C}{m}

m = molality of solute (m or \frac{mol}{Kg} )

Given : kf = 1.86 \frac{^0 C*Kg}{mol}

m = 0.907 \frac{mol}{Kg} )

Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1

Plugging value in expression :

ΔTf = 1* 1.86 \frac{^0 C*Kg}{mol} * 0.907\frac{mol}{Kg} )

ΔTf = 1.69 ° C

Hence change in freezing point = 1.69 °C

5 0
3 years ago
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