which group of the periodic table has elements with atoms that tend not to bond with other atoms of other elements

If you have 2.86 gallons of pure water, how many molecules of water are ionized at any given time ?

AleksandrR [38]

Answer:

The molar mass of water is 18 grams per mole. So in 18 grams of water, there are 6.02 x 1023 molecules.

One gallon of water is equal to 3.7854118 liters. If we assume that the water is at 4 °C (to keep things simple), the density of water is exactly 1 gram per milliliter, or 1 kilogram per liter. So 1 gallon of water at 4 °C weighs exactly 3.7854118 kilograms, or 3785.41118 grams.

If one mole of water weighs 18 grams, then there are 210.30 moles of water in a gallon. Since one mole of water is made up of 6.02 x 1023 molecules, in one gallon of water there are 1.266 x 1026 molecules.

That is: 126600000000000000000000000 molecules!

If we multiply that by the number of atoms in a single water molecule -- 3 atoms -- then we get:

379800000000000000000000000 atoms!

6 0

2 years ago

Which of the following is true about most predators? (Select all that apply.)

-Dominant- [34]

Answer:

select all of them except they are the biggest

3 0

2 years ago

Explain Newton's second law

zhenek [66]

Answer:

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object

Explanation:

5 0

3 years ago

Read 2 more answers

calculate the volume occupied by 10g of propane gas, under normal conditions of temperature and pressure

andriy [413]

Answer:

5.5 L

Explanation:

First we convert 10 g of propane gas (C₃H₈) to moles, using its molar mass:

10 g ÷ 44 g/mol = 0.23 mol

Then we use the PV=nRT formula, where:

P = 1 atm & T = 293 K (This are normal conditions of T and P)

n = 0.23 mol

R = 0.082 atm·L·mol⁻¹·K⁻¹

V = ?

1 atm * V = 0.23 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 K

V = 5.5 L

3 0

3 years ago

What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a

Vadim26 [7]

Answer:

$\boxed {\boxed {\sf 333 \ grams}}$

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

$Q= mc \Delta T$

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

ΔT= final temperature - initial temperature

The sample was heated from 58.8 degrees Celsius to 88.9 degrees Celsius.

ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

Q= 4500.0 J
c= 0.4494 J/g°C
ΔT = 30.1 °C

Substitute these values into the formula.

$4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)$

Multiply on the right side of the equation. The units of degrees Celsius cancel.

$4500.0 \ J = m (13.52694 J/g)$

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

$\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}$

The units of Joules cancel.

$\frac {4500.0 \ J }{13.52694 J/g}= m$

$332.6694729 \ g =m$

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

$333 \ g \approx m$

The mass of the sample of metal is approximately 333 grams.

8 0

2 years ago