What is the volume of 220 grams of an object with a density 55g/cm3

What kind of chemical reaction is methane + oxygen → water + carbon dioxide + heat

Veronika [31]

COMBUSTION
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:)

3 0

3 years ago

I just need answers please!!

otez555 [7]

Answer:

#1 is an organism

#2 is cell

#4 - Digestive System gets nutrients (good) from food and hands it over to the blood and Circulatory System then carries those nutrients where they need to go

4 0

3 years ago

Read 2 more answers

PLZ ANSWER QUICKLY! The electronegativity values of carbon, hydrogen, and nitrogen are compared in the table. Comparison of Elec

andriy [413]

Answer:

c. CH4 < NH3 because the NH bond is more polar than the CH bond.

Explanation:

Actually, the electronegativity difference between carbon and hydrogen is just about 0.4. This meager difference in electronegativity corresponds to a nonpolar bond between the two atoms.

However, the electronegativity difference between nitrogen and hydrogen is about 0.9. This larger electronegativity difference corresponds to the existence of a polar covalent bond between the two atoms.

Hence the N-H bond is significantly polar unlike the C-H bond. This implies that CH4 molecules are only held together by weak dispersion forces while NH3 molecules are held together by stronger dipole-dipole interactions and hydrogen bonds.

8 0

2 years ago

An insoluble base reacts with an acid to form a salt

Alex17521 [72]

Answer: you can use a strainer

Explanation:

4 0

3 years ago

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T

Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 730.5 mmHg

V = Volume of the gas = 9.49 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol$

According to the reaction shown below as:-

$2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when $\frac{2}{3}\times 0.3794$ moles of potassium chlorate undergoes reaction.

<u>Moles of potassium chlorate reacted = 0.2529 moles</u>

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 735.5 mmHg

V = Volume of the gas = 9.99 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol$

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g$

<u>Hence, the amount of oxygen gas collected will be 12.8675 g</u>

5 0

3 years ago