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vova2212 [387]
3 years ago
7

Sub atomic particle Mass

Chemistry
1 answer:
adelina 88 [10]3 years ago
6 0
The proton is 1
Electron is 1/1840
The neutron is 1
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What is the atomic number for an element whose mass number is 78,
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Answer:

protons = atomic numbermass number = protons + neutronsatomic number = mass number - neutrons = 78-38 = 40

Explanation:

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How many liters of o2 do you have if you have 5.8 moles of o2
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Answer:

130 Liters

Explanation:

if 1 mol is 22.4 L, then 5.8 mol is 130 L (129.92 but use sig figs)

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Hydrogen gas reacts with chlorine gas to form hydrogen chloride as shown in the following reaction: H2 (9) + Cl2 (g) + 2HCl (9)
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Answer:

467

Explanation:

ncl2 = 454.4x1/(71.0 g/mol) = 6.40 mols cl2

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How acidic buffer resist change in pH on addition of acid and base.
andrew-mc [135]

Answer:

Buffers are solutions that resist changes in pH, upon addition of small amounts of acid or base. The can do this because they contain an acidic component, HA, to neutralize OH- ions, and a basic component, A-, to neutralize H+ ions. Since Ka is a constant, the [H+] will depend directly on the ratio of [HA]/[A-].

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5 0
3 years ago
If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe
pishuonlain [190]

Answer:

50

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

Mᵣ:           30.01     32.00   46.01

               2NO   +   O₂ ⟶ 2NO₂

Mass/g:  80.00     16.00

2. Calculate the moles of each reactant  

\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

\text{Moles of NO}_{2} =  \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}

(b) Mass of NO reacted

\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0
3 years ago
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