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zzz [600]
3 years ago
11

If you were going to build a system to check the effectiveness of automobile catalytic converters on cars, what substances would

you want to look for in the car exhaust?
Chemistry
1 answer:
statuscvo [17]3 years ago
8 0

Answer:

Oxides of nitrogen and hydrocarbon

Explanation:

An automobile catalytic converter is a substance that catalyses redox reactions that convert toxic gases emanating from an automobile engine to less toxic gases.

The substance work by converting carbon monoxide (CO) and hydrocarbons to carbon dioxide and water and oxides of nitrogen to nitrogen and oxygen gas respectively.

<em>Hence, in order to check the effectiveness of a catalytic converter, unburned hydrocarbon and oxides of nitrogen are substances to watch out for in the car exhaust. None of these gases should be found in an exhaust with effective catalytic converter and vice versa. </em>

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If it requires 48.3 milliliters of 0.55 molar nitric acid to neutralize 15.0 milliliters of barium hydroxide, solve for the mola
Gemiola [76]
First you have to moles so multiply .0483L X .55M= .026565 Multiply moles by mole ratio which is 1/2, so the moles becomes .013283 now molarity=moles/volume; divide .013283/.015L=.885533M significant figures and you final answer is 0.89M
8 0
3 years ago
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Which type of electromagnetic wave has the greatest frequency?(1 point)
Citrus2011 [14]

c. x-rays

My answer is that x-rays or gamma rays have the greatest (or highest) frequency waves.

8 0
3 years ago
A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO
Akimi4 [234]

Answer: The empirical formula for the given compound is CH_2

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of CO_2=22.0g

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, \frac{12}{44}\times 22.0=6g of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles

Step 2: 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 0.5 moles.

For Carbon = \frac{0.5}{0.5}=1

For Hydrogen  = \frac{1.0}{0.5}=2

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is C_{1}H_{2}=CH_2

4 0
4 years ago
The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c
netineya [11]

<u>Answer:</u> The solubility of oxygen at 682 torr is 4.58\times 10^{-3}M

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{A}=K_H\times p_{A}

Or,

\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}

where,

C_1\text{ and }p_1 are the initial concentration and partial pressure of oxygen gas

C_2\text{ and }p_2 are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used:  1 atm = 760 torr

C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm

Putting values in above equation, we get:

\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M

Hence, the solubility of oxygen gas at 628 torr is 4.58\times 10^{-3}M

4 0
3 years ago
Type the formula of the following compound: Aluminum nitride
evablogger [386]

The formula for that compound is AlN

6 0
3 years ago
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