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Gwar [14]
3 years ago
10

Calculate the number of ATOMS in 1.0 mole of O2.

Chemistry
2 answers:
podryga [215]3 years ago
8 0

<u>Answer:</u> The number of atoms in given amount of oxygen gas is 6.022\times 10^{23}

<u>Explanation:</u>

According to the mole concept:

1 mole of a substance contains 6.022\times 10^{23} number of atoms.

We are given:

Moles of oxygen gas = 1 mole

So, the number of atoms that will be present in 1 mole of oxygen gas will be 1\times 6.022\times 10^{23}=6.022\times 10^{23} number of atoms.

Hence, the number of atoms in given amount of oxygen gas is 6.022\times 10^{23}

Wewaii [24]3 years ago
7 0
There is 6.02*10^23 molecule per mole. And there is 2 atoms per oxygen molecule. So the answer is 1.204*10^24 atoms in 1.0 mole of O2.
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dimaraw [331]

Answer:

neon gas

Explanation:

neon is a noble gas, meaning it has a full outer shell of elections. this means it is stable and is very unlikely to have a reaction with another substance.

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2 years ago
Calculate the tempature of a 0.50 mol sample of a gas at a pressure of 0.987 atm and a volume of 12 L
dalvyx [7]

Answer:

The temperature is 288, 88K

Explanation:

We use the formula PV= nRT:

T= PV/nR

T= 0,987 atm x12 L/0,50 mol x 0,082 l atm/K mol

<em>T=288,88K</em>

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3 years ago
What would you predict for the lengths of the bonds in no2− relative to n−o single bonds and double bonds?
djverab [1.8K]

The bonds in no2− relative to n−o

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Write the net ionic equation for any precipitation reaction that may be predicted when aqueous solutions of calcium iodide and c
Hitman42 [59]

<u>Answer:</u> The net ionic equation is written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of calcium iodide and chromium (III) sulfate is given as:

3CaI_2(aq.)+Cr_2(SO_4)_3(aq.)\rightarrow 3CaSO_4(s)+2CrI_3(aq.)

Ionic form of the above equation follows:

3Ca^{2+}(aq.)+6I^{-}(aq.)+2Cr^{3+}(aq.)+3SO_4^{2-}(aq.)\rightarrow 3CaSO_4(s)+2Cr^{3+}(aq.)+6I^-(aq.)

As, chromium and iodide ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

3Ca^{2+}(aq.)+3SO_4^{2-}(aq.)\rightarrow 3CaSO_4(s)

Hence, the net ionic equation is written above.

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