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Elanso [62]
3 years ago
8

The combining of hydrogen and oxygen gas into liquid water is an example of what kind of change?

Chemistry
1 answer:
KatRina [158]3 years ago
7 0
The combining of hydrogen and oxygen gas into liquid=H2O (water) is a compound or a molecule because it is 2 different elements combined with chemical bonds so its Chemical. B 
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1 - 2Na + O2 ➔ Na2O Balanced Not Balanced
Mrrafil [7]

Answer:

1 not

2 not

3 not

4 balanced

5 not

6 not

7 balanced

Explanation:

the amount of elements must be equal in the reactant and products

5 0
3 years ago
Question 8<br> Review<br> Which metal jis most easily oxidized?
tester [92]

Answer:

The order of some common metals in the electromotive series, starting with the most easily oxidized, is: lithium, potassium, calcium, sodium, magnesium, aluminum, zinc, chromium, iron, cobalt, nickel, lead, hydrogen, copper, mercury, silver, platinum, and gold.

Explanation:

8 0
3 years ago
Read 2 more answers
If an old car burns 0.05 mL oil every mile, How much oil will it burn if driven 100,000 miles? Show your answer in mL and L?.
Irina18 [472]

0.05 * 100000

= 5000ml

= 5L

7 0
3 years ago
A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the
Viefleur [7K]

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.

We can calculate the total number of moles using the ideal gas equation.

P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V

At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.

We can calculate the moles of nitrogen using the ideal gas equation.

P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V

The mole fraction of nitrogen in the mixture is:

X(N_2) = \frac{0.030 mol/L \times V}{0.091 mol/L \times V} = 0.33

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

Learn more: brainly.com/question/2060778

5 0
3 years ago
The fact that HBO2, a reactive compound, was produced rather than the relatively inert B2O3 was a factor in the discontinuation
Reil [10]

Answer:

1027.62 g

Explanation:

For B_2H_6  :-

Mass of B_2H_6  = 296.1 g

Molar mass of B_2H_6  = 27.66 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{296.1\ g}{27.66\ g/mol}

Moles\ of\ B_2H_6= 10.705\ mol

From the balanced reaction:-

B_2H_6(g) + 3 O2_{(l)}\rightarrow 2 HBO_2_{(g)}+ 2 H_2O_{(l)}

1 mole of B_2H_6 react with 3 moles of oxygen

Thus,

10.705 mole of B_2H_6 react with 3*10.705 moles of oxygen

Moles of oxygen = 32.115 moles

Molar mass of oxygen gas = 31.998 g/mol

<u>Mass = Moles * Molar mass = 32.115 * 31.998 g = 1027.62 g</u>

8 0
3 years ago
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