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3 years ago

Combustion reactions are characterized by the reaction of a substance with

2 answers:

8 0

Answer: Combustion reactions are characterized by the reaction of a substance with O2 to form one or more oxygen, often water. Combustion reactions emit heat, and they are important because most of society's energy is derived from them. Ex.: CH4 (g) + 2O2 (g) -- CO2 (g) + 2H2O (g)

Explanation: I hope thus helps UwU

makvit [3.9K]3 years ago

8 0

^ They are CORRECT.
Just looked over it

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Explain how things that plants or animals do are done by the work of cells

madam [21]

Animals and plants use their cells by growing and observing food and oxygen

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4 years ago

Read 2 more answers

As illustrated, the below manometer consists of a gas vessel and an open-ended U-tube containing a nonvolatile liquid with a den

STALIN [3.7K]

Answer:

1.01atm is the pressure of the gas

Explanation:

The difference in heights in the two sides is because of the difference in pressure of the enclosed gas and the atmospheric pressure. This difference is in mm of the nonvolatile liquid. The difference in mm Hg is:

32.3mm * (0.993g/mL / 13.6g/mL) = 2.36mmHg

As atmospheric pressure is 765mm Hg and assuming the gas has more pressure than the atmospheric pressure (There is no illustration), the pressure of the gas is:

765mm Hg + 2.36mm Hg = 767.36 mmHg

In atm:

767.36 mmHg * (1atm / 760 mmHg) =

1.01atm is the pressure of the gas

5 0

3 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 7.82 g of ethane is

NeTakaya

Answer:

4.79 g of water

Explanation:

From the reaction equation;

C2H6(g) + 7/2O2(g) ----> 2CO2(g) + 3H2O(g)

Next we convert the given masses of reactants to moles of reactants.

For ethane; number of moles = mass/molar mass= 7.82g/ 30gmol-1= 0.261 moles

For oxygen; number of moles= 9.9 g/32gmol-1= 0.31 moles

Next we determine the limiting reactant, the limiting reactant yields the least amount of product.

For ethane;

From the reaction equation,

1 mole of ethane yields 3 moles of water

0.261 moles of Ethan yields 0.261 ×3 = 0.783 moles of water

For oxygen;

3.5 moles of oxygen yields 3 moles of water

0.31 moles of oxygen yields 0.31 × 3/3.5 = 0.266 moles of water

Hence oxygen is the limiting reactant.

Mass of water produced = 0.266 moles of water × 18gmol-1 = 4.79 g of water

4 0

3 years ago

Suppose you start with a solution of red dye #40 that is 2.3 ✕ 10−5 M. If you do three successive volumetric dilutions pipetting

larisa [96]

Answer: Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

Explanation:

According to the neutralization law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $2.3\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$2.3\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.28\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.28\times 10^{-5}M$

2) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.28\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.28\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=0.034\times 10^{-5}M$

Therefore, the concentration of diluted solution is $0.034\times 10^{-5}M$

3) On further dilution

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = $0.034\times 10^{-5}M$

$V_1$ = volume of stock solution = 3.00 ml

$M_2$ = molarity of diluted solution = ?

$V_2$ = volume of diluted solution = 25.00 ml

$0.034\times 10^{-5}M\times 3.00=M_2\times 25.00$

$M_2=4.0\times 10^{-8}M$

Therefore, the concentration of final solution is $4.0\times 10^{-8}M$

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3 years ago

Which solid is the most suitable in the space elevator cable

Tresset [83]

It should be Carbon Nanotubes

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3 years ago