The number of atoms in a molecule varies depending on the specific molecule formed. As a rule of thumb, a molecule must be composed of at least two atoms with no maximum amount of atoms as long as all the atoms in the molecule are bonded together somehow.

8 0

4 years ago

Calculate the amount of water (in grams) that must be added to (a) 6.80 g of urea [(NH2)2CO] in the preparation of a 9.95 percen

Lemur [1.5K]

Explanation:

Amount of water required in each case:

(a)The mass% of the solution is:9.95

Mass of solute that is urea is 6.80 g

To determine the mass of solvent water use the formula:

$mass percent=\frac{mass of solute}{mass of solution} x 100\\\\9.95=(6.80g/mass of solution )x100\\mass of solution =(6.80 /9.95)x100\\=68.3 g$

Hence the mass of solvent = mass of solution - the mass of solute

=68.3 g - 6.80g

=61.5 g

Hence, the answer is mass of solvent water required is 61.5 g.

(b) Given mass%=1.70

mass of solute MgBr2 = 29.3 g

The mass of solvent water required can be calculated as shown below:

$mass percent=\frac{mass of solute}{mass of solution} x 100\\\\1.70=(29.3g/mass of solution )x100\\mass of solution =(29.3 g /1.70)x100\\=1720 g$

The mass of the solution is 1720 g.

Mass of solvent water = mass of solution - mass of solute

=1720 g - 29.3 g

=1690.7 g

Answer: The mass of water required is 1690.7 g.

5 0

3 years ago

A 75.0-milliliter lightbulb is filled with neon. There are 7.16 × 10-4 moles of gas in it, and the absolute pressure is 116.8 ki

Alexeev081 [22]

n = number of mole of sample of helium gas = 7.16 x 10⁻⁴ moles

T = temperature of the gas = temperature of the bulb ?

V = Volume of helium gas = 75 mL = 7.5 x 10⁻⁵ m³

P = pressure of the helium gas = 116.8 kPa = 116.8 x 10³ Pa (1 k = 1000)

Using the equation

P V = n R T

inserting the values

(116.8 x 10³) (7.5 x 10⁻⁵) = (7.16 x 10⁻⁴) (8.314) T

T = 1471.6 C

8 0

3 years ago

Read 2 more answers

If 6.5 mol NO2 react with 1.0 mol H20, how many moles of the excess reactant

harkovskaia [24]

Answer:

3.5 moles of NO2.

Explanation:

The balanced equation for the reaction is given below:

3NO2 + H2O —> NO + 2HNO3

Next, we shall determine the excess reactant. This is illustrated below:

From the balanced equation above,

3 moles of NO2 reacted with 1 moles H2O.

Therefore, 6.5 moles of NO2 will react with = (6.5 x 1)/3 = 2.17 moles of H2O.

From the above calculations, we can see that it will take a higher amount i.e 2.17 moles than what was given i.e 1 mole of H2O to react completely with 6.5 moles of NO2.

Therefore, NO2 is the excess reactant and H2O is the limiting reactant.

Next, we shall determine the number of mole of the excess reactant that reacted in the reaction. This is illustrated below:

From the balanced equation above, we can see that 3 moles of NO2 reacted with 1 mole of H2O.

Finally, we shall determine the leftover mole of the excess reactant after the reaction. This is illustrated below:

Given mole of NO2 = 6.5 moles

Reacted mole of NO2 = 3 moles

Leftover mole of NO2 =...?

Leftover mole = Given mole – Reacted mole

Leftover mole of NO2 = 6.5 – 3

Leftover mole of NO2 = 3.5 moles.

Therefore, 3.5 moles of NO2 remained after the reaction.

3 0

4 years ago