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dlinn [17]
3 years ago
14

Water is formed when two hydrogens bond to a oxygen atom

Chemistry
1 answer:
tangare [24]3 years ago
6 0

Answer:

c) a compound

Explanation:

a compound is something made up of two or more separate elements, the separate elements in water are hydrogen(H) and oxygen(O)

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Calculate the oxidation number of the chlorine in perchloric acid, hclo4, a strong oxidizing agent.
Sveta_85 [38]
<span>The oxidation number that is given to chlorine in a chemical combination indicates how many electrons are gained or lost because it is a neutral compound, perchloric acid has a total charge of 0. Hydrogen has an oxidation number of +1,whereas oxygen has an oxidation number of -2. So for the total charge to be 0, 1 hydrogen + 4 oxygen of negative charge added together must equal 0, which will make the oxidation number of chlorine, solving this algebraically +7.</span>
7 0
4 years ago
What type of bond occurs when two atoms share one pair of valence electrons?
mestny [16]

A covalent bond occurs when two atoms share one pair of valence electrons

Brainliest answer please? :)

5 0
3 years ago
Read 2 more answers
Chemistry help will be grateful thanks
igor_vitrenko [27]

Answer:

a = double displacement(iii)

b = Single displacement (ii)

c = Synthesis reaction (i)

d = decomposition reaction (iv)

Explanation:

It's not that hard!

6 0
3 years ago
How can weak acids and weak bases cause pH stress to cells?
telo118 [61]

Answer:

b

Explanation:

If a cell has transporters that are not well regulated, weak acids or weak bases can accumulate inside the cells and cause stress.

3 0
3 years ago
Using the standard enthalpies of formation for the chemicals involved, calculate the enthalpy change for the following reaction.
Firlakuza [10]

Answer: -134 kJ

Explanation:

The balanced chemical reaction is,

3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq)+NO(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{HNO_3}\times \Delta H_{HNO_3})+(n_{NO}\times \Delta H_{NO})]-[(n_{H_2O}\times \Delta H_{H_2O})+(n_{NO_2}\times \Delta H_{NO_2})]

where,

n = number of moles

Now put all the given values in this expression, we get

\Delta H=[(2\times -207)+(1\times 90)]-[(1\times -286)+(3\times 32)]

\Delta H=-134kJ

Therefore, the enthalpy change for this reaction is, -134 kJ

6 0
3 years ago
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