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kumpel [21]
3 years ago
7

Which element is always present in a combustion reaction?

Chemistry
2 answers:
ikadub [295]3 years ago
6 0
I think the answer is oxygen is correct.
diamong [38]3 years ago
4 0
The element <u>Oxygen</u> is always present in a combustion reaction.
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Match each of the three descriptions of a volume to the appropriate metric unit of volume
tankabanditka [31]
A.) 1.
b.) 3
c.) 2
I hope this helps.
8 0
3 years ago
Nickel + oxygen = nickel oxide. What is the balanced redox reaction?
uranmaximum [27]

Answer:

This is an oxidation-reduction (redox) reaction:

2 Ni0 - 4 e- → 2 NiII

(oxidation)

2 O0 + 4 e- → 2 O-II

(reduction)

Ni is a reducing agent, O2 is an oxidizing agent.

7 0
2 years ago
Help me with this work plz plz
storchak [24]

Answer:

a -4

b - 8

c - 5

d 2

Explanation:

Significant Figures: The number of digits used to express a measured or calculated quantity. By using significant figures, we can show how precise a number is. ... Accuracy: Refers to how closely individual measurements agree with the correct or true value.

7 0
2 years ago
If we have 1.23 mol of NaOH in solution and 0.85 mol of Cl2 gas is available to react, which one is the limiting reactant? Give
Vinil7 [7]

Answer:

NaOH is the limiting reactant.

Explanation:

Hello there!

In this case, since the reaction taking place between sodium hydroxide and chlorine has is:

NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O

Which must be balanced according to the law of conservation of mass:

2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O

Whereas there is a 2:1 mole ratio of NaOH to Cl2, which means that the moles of the former that are consumed by 0.85 moles of the latter are:

n_{NaOH}=0.85molCl_2*\frac{2molNaOH}{1molCl_2}\\\\n_{ NaOH}=1.7molNaOH

Therefore, since we just have 1.23 moles out of 1.70 moles of NaOH, we infer this is the limiting reactant.

Regards!

3 0
3 years ago
8.
White raven [17]

Answer:

mixture of four compounds

mixture of two elements and two compounds

Explanation:

  1. it can be one of those
  2. can u tell me if its correct pls I'm studying this too

6 0
3 years ago
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