What is the change of the oxidation number arsenic in the following reaction

How many H2 and O2 molecules do you need to make exactly 4 water molecules with no leftovers?

vlabodo [156]

You need 4 H2 molecules and 2 O2 molecules.
This is because water is H2O which is 2 Hydrogen and 1 Oxygen so if you want to make 4 water molecules, you’ll need 8 hydrogen and 4 oxygen.
Given H2 you need 4 because 4•2 is 8 and for O2 you need 2 because 2•2 is 4.
I hope this makes sense.

3 0

3 years ago

How many moles is 55 g of Naci? 0.025 g of NaCO3?

mart [117]

Answer:

0.940mol &

0.000301mol respectively.

Explanation:

number of moles = given mass / molar mass

given mass of Nacl = 55g Molar mass = 23 + 35.5

n=m/M = 55g/58.5g/mol = 0.940mol

note- (add the atomic weights of sodium and chlorine to get the molar mass of Nacl.) = 58.5g/mol

similarly, NaCO3 = 23 + 12 + 16*3 = 83g/mol

n=m/M = 0.025g/83g/mol = 3.01 * 10^-4 = 0.000301mol

extra: If you ever get asked to put it in number of particles just use the relation of 1mole = 6.02 * 10^23 particles.

8 0

3 years ago

Lakshmi has a sample of ammonium nitrate (NH4NO3) that has a mass of 40.10 g. She knows that the molar mass of NH4NO3 is 80.0432

avanturin [10]

The answer is 0.5010

Number of moles (n) is equal to the quotient of mass (m) and molar mass of a sample (Mr):
n = m/Mr

We have:
n = ?
m = 40.10 g
Mr = 80.0432 g/mol

n = 40.10 g : 80.0432 g/mol = 0.5010

40.10 has 4 significant digits,
80.0432 has 6 significant digits.
Since 4 is less than 6, we choose 4 significant digits

5 0

3 years ago

Consider the reaction below in a closed flask. At 200 o C, the equilibrium constant (Kp) is 2.40 × 103 . 2 NO (g)  N2 (g) + O2

olga55 [171]

Explanation:

Since, the given reaction is as follows.

$2NO(g) \rightleftharpoons N_{2}(g) + O_{2}(g)$

Initial: 36.1 atm 0 0

Change: 2x x x

Equilibrium: (36.1 - 2x) x x

Now, expression for $K_{p}$ of this reaction is as follows.

$K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}$

As the initial pressure of NO is 36.1 atm. Hence, partial pressure of $O_{2}$ at equilibrium will be calculated as follows.

$K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}$

$2.40 \times 10^{3} = \frac{x \times x}{(36.1 - 2x)^{2}}$

x = 18.1 atm

Thus, we can conclude that partial pressure of $O_{2}$ at equilibrium is 18.1 atm.

5 0

3 years ago

Read 2 more answers

Suppose 0.410 kg of hexane are burned in air at a pressure of exactly 1 atm and a temperature of 13.0 °C. Calculate the volume o

3241004551 [841]

Answer:

The answer is 671 litres of carbon dioxide is produced from 0.410 kg of hexane

Explanation:

We first write a balanced reaction for the complete combustion of hexane thus

The stoichiometry of the cumbustion of hexane in air is

2C6H14(g)+18O2(g)→12CO2(g)+14H2O(l) or

C6H14(g)+9O2(g)→6CO2(g)+7H2O(l)

From the above reaction it is observed that one mole of hexane burns completely in the presence of oxygen to produce 6 moles of carbon dioxide

Therefore we calculate the nuber of moles of hexane present in the sample thus

Mass hexane of sample = 0.41 kg

Molar nass of hexane = 86.18 g/mol

number of moles of hexane = (mass of hexane)/(molar mass of hexane) = (0.41×1000)/86.16 = 410/86.16 = 4.76 moles

As we have seen from the chemical reaction, 1 mole of H6H14 produces 6 moles of CO2 hence 4.76 moles of Hexane produces

4.76×6 moles of CO2 which is 28.55 moles of CO2

From the question we have the temperature and the pressure of the production of CO2 as

Temperature of reaction = 13° C converting to kelving gives= 13+273.15 = 286.15 K

and pressure = 1 atmosphere or 101325 Pa

13.0∘C=13.0∘C+273.15=286.15 K

The volume of the produced CO2 can be calculated using the combined ideal gas equation given by

P×V=n×R×T where

Here

P = Gas pressure (of CO2 )

V = Volume (of the CO2)

n = number of moles of gas (CO2) present

R = universal gas constant, equal to 0.0821 atm× L/(mol× K )

T = absolute temperature in Kelvin

Thus we have

1×V = 28.55×0.0821×286.15 or V = 670.76L

Rounding up the answer to 3 significant digits we have

670.76L ≅ 671L

671 litres of carbon dioxide is produced from 0.410 kg of hexane

8 0

4 years ago