How is nitric acid classified?

if a balloon had an initial volume of 100ml at pressure of 3.2 atm and the pressure was decreased to 1.9 atm, what would the new

hram777 [196]

Answer:

168.4 mL

Explanation:

Data Given

initial volume V1 of gas in balloon = 100 mL

initial pressure P1 of gas in balloon = 3.2 atm

final pressure P2 of gas in balloon = 1.9 atm

final volume V2 of gas in balloon = ?

Solution:

This problem will be solved by using Boyle's law equation at constant Temperature.

The formula used

P1V1 = P2V2

As we have to find out Volume, so rearrange the above equation

V2 = P1V1 / P2

Put value from the data given

V2 = 100 mL x 3.2 atm / 1.9 atm

V2 = 168.4 mL

So the final Volume of gas in baloon = 168.4 mL

4 0

3 years ago

What are the products of the balanced equation for the combustion of C8H17OH ?

erastova [34]

C8H17OH + 12O2 —> 8CO2 + 9H2O

3 0

3 years ago

The starting substanceses of a chemical reaction are refered to as

Debora [2.8K]

The starting substances in a chemical reaction are called reactants - they are written on the left side of a chemical equation.

6 0

3 years ago

The solubility of nitrogen gas at 25 ◦C and 1 atm is 6.8×10−4 mol/L. If the partial pressure of nitrogen gas in air is 0.76 atm,

iris [78.8K]

Answer:

Concentration of dissolved nitrogen = 5.2 × 10⁻⁴ mol/L

Explanation:

More the pressure of the gas, more will be its solubility.

So, for two different pressure, the relation between them is shown below as:-

$\frac {P_1}{C_1}=\frac {P_2}{C_2}$

Given ,

P₁ = 1 atm

P₂ = 0.76 atm

C₁ = 6.8 × 10⁻⁴ mol/L

C₂ = ?

Using above equation as:

$\frac{1\ atm}{6.8\times 10^{-4}\ mol/L}=\frac{0.76\ atm}{C_2}$

$C_2=\frac{0.76\times 6.8\times 10^{-4}}{1}\ mol/L$

Concentration of dissolved nitrogen = 5.2 × 10⁻⁴ mol/L

5 0

3 years ago

100 PIONTSSSSS HELP ASAP

valina [46]

Left Panel

Short answer A

Solution

Since you have been given choices, my sloppy numbers will do, but it anyone is going to see this, YOU SHOULD CLEAN THEM UP WITH THE NUMBERS THAT COME FROM YOUR PERIODIC TABLE.

Equation

Sodium Phosphate + Calcium Chloride ===> Sodium Chloride + Calcium Phosphate.

Na3PO4 + CaCl2 ===> NaCl + Ca3(PO4)2

Step One

Balance the Equation

2Na2PO4 + 3CaCl2 ==> 6NaCl + Ca3(PO4)2

Step Two

Find the molar mass of CaCl2

Ca = 40

2Cl = 71

Molar Mass = 40 + 71 = 111 grams/mole

Step Three

Find the number of moles of CaCl2

Given mass = 379.4

Molar Mass = 111

moles = given Mass / molar Mass

moles of CaCl2 = 379.4/111 = 3.418 moles

Step Four

Find the number of moles of Ca3(PO4)2 needed.

This requires that you use the balance numbers from the balanced equation.

For every 3 moles of CaCl2 you have, you get 1 mole of Ca3(PO4)2

n_moles of Ca3(PO4)2 = 3.418 / 3 = 1.13933 moles

Step Five

Find the molar mass of Ca3(PO4)2

From the periodic table,

3Ca = 3 * 40 = 120

2 P = 2 * 31 = 62

8 O = 8 * 16 =128

Molar Mass = 120 + 62 + 128= 310 grams per mole.

Step Six

1 mole of Ca3(PO4)2 has a molar mass of 310 gram

1.13933 moles of Ca3(PO4)2 = x

x = 1.13933 moles * 310 grams /mole

x = 353.2 grams. As you can see, even with my rounding I'm only out 0.3 of a gram. DON'T FORGET TO PUT THIS TO THE PROPER SIG DIGS IF SOMEONE ELSE IS GOING TO SEE IT.

Middle Panel

Short Answer C

Equation

2HCl + Mg ===> H2 + MgCl2

The object of the first part of the game is to find the number of moles of H2.

Step One

Find the moles of HCl

1 mole HCl = 35.5 + 1 = 36.5

n = given mass divided by molar mass

n = 49 grams / 36.5 = 1.34 moles.

The balanced equation tells you that for ever mole of H2 produced, you need 2 moles of HCl. That's what the balance numbers are for.

So the number of moles of H2 is 1.34 / 2 = 0.671 moles of H2.

Now we come to Part II. We have to use an new friend of yours that I have seen only once before from you.

Find V using PV = nRT

R is going to be in kPa so the value of R = 8.314

V = ???

n = 0.671 moles

T = 25 + 273 = 298oK

P = 101.3 kPa

101.3 * V= 0.671*8.314 * 298

V = 0.671 * 8.314 * 298 / 101.3

V = 16.4

The answer is C and again, I have rounded almost everything except R, although it can go out to 8 places.

Right Panel

I can't see the panel. I don't know what the problem is. Never mind I got it. I'm going to be a little skimpy on this one since I've done two like it and they are long.

LiOH + HBr ===> LiBr + H2O and the equation is balanced.

You have to figure out the moles of LiOH and HBr. Use the LOWEST number of moles

n_LiOH = given mass / molar mass = 117/(7 + 16 + 1) = 117 / 24 = 4.875 moles

n_HBr = given mass / molar mass = 141/(1 + 80) = 141 / 81 = 1.741 moles

HBr is the lower number. That's all the LiBr you are going to get is 1.741. There is no adjustment to be made from the balance equation.

n = given mass / molar mass multiply both sides by the molar mass

n * Molar mass (LiBr) = n * (7 + 80) = 1.741 * 87 = 151 grams of

The answer is C

6 0

3 years ago