Is ethane polar or nonpolar

What is the Scientific notation for 8,950,000

Tatiana [17]

The scientific notation for 8,950,000 is
8.95 × 10^6

6 0

3 years ago

In order for the process to be endothermic, the energy required to break the lattice has to be _______ than the energy released

densk [106]

Explanation:

I honestly believe the answer is greater

8 0

3 years ago

Read 2 more answers

Matt is conducting an experiment where he compares the properties of water and

NikAS [45]

Answer:

sitric acid from lemon

Explanation:

there

3 0

3 years ago

What is the volume of 14.0g of nitrogen gas at STP?

lozanna [386]

Answer:

The volume of 14.0 g of nitrogen gas at STP is 11.2 liter.

Explanation:

STP stands for standard pressure and temperature.

The International Institute of of Pure and Applied Chemistry, IUPAC changed the definition of standard temperature and pressure (STP) in 1982:

Before the change, STP was defined as a temperature of 273.15 K and an absolute pressure of exactly 1 atm (101.325 kPa).

After the change, STP is defined as a temperature of 273.15 K and an absolute pressure of exactly 105 Pa (100 kPa, 1 bar).

Using the ideal gas equation of state, PV = nRT you can calculate the volume of one mole (n = 1) of gas. With the former definition, the volume of a mol of gas at STP, rounded to 3 significant figures, was 22.4 liter. This is classical well known result.

With the later definition, the volume of a mol of gas at STP is 22.7 liter.

I will use the traditional measure of 22.4 liter per mole of gas.

1) Convert 14.0 g of nitrogen gas to number of moles:

n = mass in grams / molar mass

Atomic mass of nitrogen: 14.0 g/mol

Nitrogen gas is a diatomic molecule, so the molar mass of nitrogen gas = molar mass of N₂ = 14.0 × 2 g/mol = 28.0 g/mol

n = 14.0 g / 28.0 g/mol = 0.500 mol

2) Set a proportion to calculate the volume of nitrogen gas:

22.4 liter / mol = x / 0.500 mol

Solve for x: x = 0.500 mol × 22.4 liter / mol = 11.2 liter.

Conclusion: the volume of 14.0 g of nitrogen gas at STP is 11.2 liter.

6 0

3 years ago

You have a 2.0 mL sample of acetic acid (molar mass 60.05 g/mol) of unknown concentration. You titrate it to its endpoint with 2

Katyanochek1 [597]

Answer: The mass of acetic acid used is 0.12 grams

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $CH_3COOH$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=2.0mL\\n_2=1\\M_2=0.1M\\V_2=20.0mL$

Putting values in above equation, we get:

$1\times M_1\times 2.0=1\times 0.1\times 20.0\\\\M_1=\frac{1\times 0.1\times 20.0}{1\times 2.0}=1M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of acetic acid = ? g

Molar mass of acetic acid = 60.05 g/mol

Molarity of solution = 1 M

Volume of the solution = 2.0 mL

Putting values in above equation, we get:

$1mol/L=\frac{\text{Mass of acetic acid}\times 1000}{60.05g/mol\times 2.0}\\\\\text{Mass of acetic acid}=\frac{1\times 60.05\times 2}{1000}=0.12g$

Hence, the mass of acetic acid used is 0.12 grams

5 0

3 years ago