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3 years ago

What type of pollution in cities composed of car exhaust and industrial pollutes

Chemistry

2 answers:

stich3 [128]3 years ago

5 0

Are there any options. because that would be air pollution but is there specific terms?

Troyanec [42]3 years ago

4 0

Air pollution is the answer

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3.Write the chemical equation for the reaction when methane burns in (2)

IceJOKER [234]

Answer:

a)CH4 + O2 = CO2 + H2O

b)4CH4+5O=2CO+8H2O+2C

5 0

3 years ago

Liquid nitrogen has a density of 0.807 g/ml at –195.8 °c. if 1.00 l of n2(l) is allowed to warm to 25°c at a pressure of 1.00 at

Flauer [41]

Step 1: Change density from g/mL to g/L;

0.807 g/mL = 807 g/L

Step 2: Find Moles of N₂;
As,
Density = Mass / Volume
Or,
Mass = Density × Volume

Putting Values,

Mass = 807 g/L × 1 L

Mass = 807 g
Also,
Moles = Mass / M.mass

Putting values,

Moles = 807 g / 28 g.mol⁻¹

Moles = 28.82 moles

Step 3: Apply Ideal Gas Equation to Find Volume of gas occupied,

As,
P V = n R T

V = n R T / P
Putting Values, remember! don't forget to change temperatue into Kelvin (25 °C + 273 = 298 K)

V = (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm

V = 704.76 L

8 0

3 years ago

How many moles of O2 are needed to burn 2.56 moles of CH3OH?

lukranit [14]

Answer:

$n_{O_2}=3.84molO_2$

Explanation:

Hello!

In this case, since the combustion reaction of methanol is:

$CH_3OH+\frac{3}{2} O_2\rightarrow CO_2+2H_2O$

In such a way, since there is 1:3/2 mole ratio between methanol and oxygen, we can compute the moles of oxygen that are needed to burn 2.56 moles of methanol as shown below:

$n_{O_2}=2.56molCH_3OH*\frac{\frac{3}{2}molO_2}{1molCH_3OH} \\\\n_{O_2}=3.84molO_2$

Best regards!

6 0

2 years ago

he combustion of propane (C3H8) is given by the balanced chemical equation C_3H_8+5O_2\longrightarrow3CO_2+4H_2O C 3 H 8 + 5 O 2

Basile [38]

Answer:

290 grams

Explanation:

Let's begin by writing the balanced chemical equations:

$C_{3}H_{8} +5O_{2} --->3CO_{2} +4H_{2}O$

Then we calculate the number of moles in 97g of propane.

n(propane)= $\frac{mass}{molarmass} =\frac{97g}{44.1g/mol}=2.1995mol$

According to the balanced chemical equation, one mole of propane produces 3 moles of carbon dioxide. So the available number of moles of propane must be multiplied by three to work out the number of carbon dioxide produced.

n(carbon dioxide)= 2.1995mol*3 = 6.5985mol

mass(carbon dioxide) = moles * molar mass

= 6.5985 mol * 44.01 g/mol

= 290 grams

4 0

3 years ago

1. A rock with a mass of 66.5g<br>occupies 23.0 cm. Density?

telo118 [61]

Answer:

2.89 g/cm^3

Explanation:

Since density equals mass over volume (or also seen as $d=\frac{m}{v}$ ), simply divide 66.5 grams by 23.0 cm. This will output an answer of 2.89 g/cm^3.

7 0

3 years ago