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3 years ago

What volume of oxygen will react with 50cm^3 of nitrogen monoxide?

Chemistry

1 answer:

Mashcka [7]3 years ago

7 0

Answer:

GIVE BRAINLIEST PLEASE!!! Answer - 50cm^3 of oxygen

Explanation:

They need (almost) equal amounts of substance to react

hope this helps <3

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Give the formula of the conjugate base of each of the following. (Type your answer using the format CO2 for CO2, (NH4)2CO3 for (

MissTica

Answer:

a) [NH2]-,

b) [O]2-

c) [Cl]-

Explanation:

The conjugate base is the base member (X⁻) of a weak acid (HX). In other words, the conjugate base is the remaining substance due to the proton loss in the acid HX.

a) For NH₃, (HX; X: NH₂⁻), conjugate base is NH₂⁻. In the format, [NH2]-.

b) For OH⁻, (HX; X: O²⁻), conjugate base is O²⁻. In the format, [O]2-.

c) For HCl, (HX; X: Cl⁻), conjugate base is Cl⁻. In the format, [Cl]-.

7 0

3 years ago

A student was asked to separate a mixture of chemicals. The weight of each component within the mixture is as follows: potassium

hram777 [196]

Answer:

Explanation:

6 0

4 years ago

The primary gas in a volcano is: water vapor carbon dioxide sulfur dioxide nitrogen

Snezhnost [94]

Answer:

Water vapor

Explanation:

The magma consist of dissolved gases when these gases produce the force the volcanic eruption take place. The volcanic gases comes out and their volume is increased tremendously. The gases present in volcano are listed below:

The volcanic gases consist of water vapors, carbon dioxide and sulfur.

These three re the primary gases but the water is present in higher amount.

The percentage of water is 60%.

The carbon dioxide present in 10-40%.

Other gases present in valcano are nitrogen, argon, helium, neon methane and hydrogen.

7 0

3 years ago

Read 2 more answers

When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x

Leno4ka [110]

You have to use the equation PV=nRT.
P=pressure (in this case 1.89x10^3 kPa which equals 18.35677 atm)
1V=volume (in this case 685L)
n=moles (in this case the unknown)
R=gas constant (0.08206 (L atm)/(mol K))
T=temperature (in this case 621 K)
with the given information you can rewrite the ideal gas law equation as n=PV/RT.
n=(18.35677atm x 685L)/(0.08206atmL/molK x 621K)
n=246.8 moles

8 0

3 years ago

Please help me like now please

777dan777 [17]

Answer:

1-1) NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

1-2) 0.5 mole of CO2

2-1) 2C4H10 + 13O2 --> 8CO2 + 10H2O

2-2) 4 mol CO2

Explanation:

Question 1

NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

To balance the equation, count the number of atoms on both sides of the equation

(1 Na, 1+3+1H, 1+1+1C, 3+2Oxygen) --> (1 Na, 1+1+1C, 3+2H, 2+1+2Oxygen)

Combining the pluses will give you the following

(1 Na, 5H, 3C, 5Oxygen) --> (1 Na, 3C, 5H, 5Oxygen)

Both sides are the same, therefore the chemical equation is balanced (originally).

From the equation, we can see that 1 mole of NaHCO3 produces 1 mole of CO2.

So that means 0.5 mole of NaHCO3 would produce 0.5 mole of CO2.

Question 2

C4H10 + O2 --> CO2 + H2O

Again, count the number of atoms on both sides of the equation

(4C, 10H, 2O) --> (1C, 2H, 3O) This time left does not equal right side

You now need to find factors that can make both sides equal.

C4H10 + O2 --> 4CO2 + H2O Now the C is balanced, let's recount 

(4C, 10H, 2Oxygen) --> (4C, 8+1Oxygen, 2H) H&O is still not balanced

C4H10 + O2 --> 4CO2 + 5H2O Now the H is balanced, let's recount

(4C, 10H, 2Oxygen) --> (4C, 8+5Oxygen, 10H) O is still not balanced

C4H10 + (13/2)O2 --> 4CO2 + 5H2O Now the O is balanced

(4C, 10H, 13Oxygen) --> (4C, 13Oxygen, 10H)

But because 13/2 is a fraction, we want to eliminate that by multiplying every reactant and product by 2 (the denominator).

2C4H10 + 13O2 --> 8CO2 + 10H2O Now it's completely balanced!

(8C, 20H, 28Oxygen) --> (8C, 28Oxygen, 20H) Yayy! It's balanced.

Now, 2 mol C4H10 produces 8 mol CO2.

So 1 mol C4H10 produces 4 mol CO2.

6 0

3 years ago