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2 years ago

How many moles of water are in 1.23 x 10 to the 18th power water molecules

Chemistry

2 answers:

Ksivusya [100]2 years ago

6 0

Answer:

The answer is 2.04x10^-6 moles of water.

Explanation:

Avogadro's number is defined as the number of particles found in an amount of substance per mole. It is the factor that relates the moles of a substance to the mass of that substance. We will use Avogadro's number to calculate the number of moles of water. as follows:

Avogadro's number = 1 mol = 6.022 x 10^23 particles.

we will use the conversion factor to calculate the number of moles of water:

Moles of water = 1.23x10^18 particles x (1mol/6.022x10^23 particles) = 2.04x10^-6 moles

mamaluj [8]2 years ago

3 0

Divide the number of molecules you have by, 6.022 x 10^23. This will give you the moles of water, or the moles of anything, since there is always 6.022 x 10^23 molecules in 1 mole of substance.

1.23x10^24 atoms/6.022x10^23 atom/mole = 2.04 mole H20

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A 1.00 liter container holds a mixture of 0.52 mg of He and 2.05 mg of Ne at 25oC. Determine the partial pressures of He and Ne

Ymorist [56]

Answer:

pHe = 3.2 × 10⁻³ atm

pNe = 2.5 × 10⁻³ atm

P = 5.7 × 10⁻³ atm

Explanation:

Given data

Volume = 1.00 L

Temperature = 25°C + 273 = 298 K

mHe = 0.52 mg = 0.52 × 10⁻³ g

mNe = 2.05 mg = 2.05 × 10⁻³ g

The molar mass of He is 4.00 g/mol. The moles of He are:

0.52 × 10⁻³ g × (1 mol / 4.00 g) = 1.3 × 10⁻⁴ mol

We can find the partial pressure of He using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 3.2 × 10⁻³ atm

The molar mass of Ne is 20.18 g/mol. The moles of Ne are:

2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol

We can find the partial pressure of Ne using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 2.5 × 10⁻³ atm

The total pressure is the sum of the partial pressures.

P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm

6 0

2 years ago

Write an equation to show how HC2O4− can act as a base with HS− acting as an acid.

Artyom0805 [142]

An acid donates $H^{+}$ ion in aqueous solution. A base accepts $H^{+}$ ion in aqueous solution.

The equation representing the acid base reaction of $HC_{2}O_{4}^{-}$ and $HS^{-}$ :

$HC_{2}O_{4}^{-}(aq) + HS^{-}(aq) ----> H_{2}C_{2}O_{4}(aq)+S^{2-}(aq)$

In the above reaction, as $HC_{2}O_{4}^{-}$ acts as a base it is accepting the hydrogen ion from $HS^{-}$ . Similarly, $HS^{-}$ donates its hydrogen ion to $HC_{2}O_{4}^{-}$ acting as an acid.

7 0

2 years ago

The list below includes some of the properties of butane, a common fuel. Identify the chemical properties in the list. Check all

Brilliant_brown [7]

<u>Full Question:</u>

The list below includes some of the properties of butane, a common fuel. Identify the chemical properties in the list. Check all of the boxes that apply.

denser than water

burns readily in air

boiling point of –1.1°C

odorless

does not react with water

burns readily in air

does not react with water

<h3><u>Explanation:</u></h3>

The type of alkane that is used in many products includes Butane. It is found as a natural gas in the environment. It is found on the deeper part of ground. It can be obtained by drilling process and gets used up in many of the products that is used for commercial purposes.

Molecular mass that is associated with butane is 58.12 g/mol. The boiling point of butane is -1 degree Celsius and -140 degree Celsius is its melting point. It is a liquefied gas and does not react with water. It will burn in air more readily.

6 0

2 years ago

James was given a sample of sodium bicarbonate NaHCO3 in a weighing dish to analyze. His teacher asked him to find the percent o

gtnhenbr [62]

This question could be answered easily if the results of the abundance of the other elements are given. You will just have to subtract the sum of all their abundances to 100. Since it's not given, the solution would just be:

Na = 23 g/mol* 1 = 23 g
H = 1 g/mol * 1 = 1 g
C = 12 g/mol * 1 = 12 g
O = 16 g/mol * 3 = 48 g
Total = 84 g

% O = 48/84 * 100 = <em>57.14%</em>

5 0

2 years ago

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In the reaction N2 + 3H2 ⇌ 2NH3, an experiment finds equilibrium concentrations of [N2] = 0.1 M, [H2] = 0.05 M, and [NH3] = 0.00

jasenka [17]

The equilibrium constant Kc for this reaction is calculated as follows

from the equation N2 + 3H2 =2 NH3

qc = (NH3)2/{(N2)(H2)^3}

Qc is therefore = ( 0.001)2 /{(0.1) (0.05)^3} = 0.08

3 0

2 years ago

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