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Oksana_A [137]
3 years ago
10

How to write 0.99966788 in scientific notation?

Chemistry
2 answers:
inn [45]3 years ago
7 0

Answer:

9.9966788 x 10^-1

mars1129 [50]3 years ago
4 0
Move the decimal 4 times to the right than thay number tomes 10^-4
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If 0.251moles of H2O gas are produced, how many liters of oxygen gas was used?
Alborosie

Answer:

o.251 prduces 45.7L of oxogen

Explanation:

hope this helps

4 0
4 years ago
What else is produced during the combustion of butane, C4H10?<br><br> 2C4H10 + 13O2 → <br> + 10H2O
Neko [114]
<h3>Answer:</h3>

8CO₂

<h3>Explanation:</h3>

We are given;

  • Butane, C₄H₁₀
  • Butane is a hydrocarbon in the homologous series known as alkane.

We are required to determine the other product produced in the combustion of butane apart from water.

  • We know that the complete combustion of alkane yields carbon dioxide and water.
  • Therefore, combustion of butane will yield carbon dioxide and water.
  • The balanced equation for the complete combustion of butane will be;

       2C₄H₁₀ + 13O₂ →  8CO₂ + 10H₂O

8 0
3 years ago
Read 2 more answers
Which can associate a suspect and the firing of a gun?
baherus [9]
D. powder residues.
the police and forensic chemists usually perform a qualitative test called GPR or gun powder residue. the residue sticks to the skin.
5 0
3 years ago
The density of a gas cannot be measured.<br> True<br> False
likoan [24]

Answer:

False

Explanation:

7 0
3 years ago
What is the excess reactant in the combustion of 23 g of methane in the open atmosphere?
Marta_Voda [28]

Answer : The excess reactant in the combustion of methane in opem atmosphere is O_{2} molecule.

Solution : Given,

Mass of methane = 23 g

Molar mass of methane = 16.04 g/mole

The Net balanced chemical reaction for combustion of methane is,

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

First we have to calculate the moles of methane.

\text{ Moles of methane}=\frac{\text{ Given mass of methane}}{\text{ Molar mass of methane}} = \frac{23g}{16.04g/mole} = 1.434 moles

From the above chemical reaction, we conclude that

1 mole of methane react with the 2 moles of oxygen

and 1.434 moles of methane react to give \frac{2moles\times 1.434moles}{1moles} moles of oxygen

The Moles of oxygen = 2.868 moles

Now we conclude that the moles of oxygen are more than the moles of methane.

Therefore, the excess reactant in the combustion of methane in open atmosphere is O_{2} molecule.


6 0
3 years ago
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