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tamaranim1 [39]
2 years ago
14

Plssss help i am giving out a lot of points

Chemistry
1 answer:
Kruka [31]2 years ago
7 0
Think the answer is zinc + hydrochloride acid zinc choride and hydrogen gasZn + 2HCI - ZnCI2 + H2 sorry I took so long I was trying to get this answer hope this helps.
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What is the period number in which helium is found
julsineya [31]

Answer:

Helium is the second element on the periodic table. It is located in period 1 and group 18 or 8A on the righthand side of the table. This group contains the noble gases, which are the most chemically inert elements on the periodic table. Each He atom has two protons and usually two neutrons and two electrons.

Explanation:

thoughtco.com

6 0
3 years ago
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What is the name of the salt with the structure K2S
gavmur [86]
Potassium sulphide :)
6 0
3 years ago
Which of the following statements are true? (There are 2)
Shalnov [3]

Answer:

A and B

Explanation:

the other two make no sense at all

4 0
2 years ago
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What is the mass in grams of 5.00moles of CH4?
anastassius [24]

Answer:

1. 80g

2. 1.188mole

Explanation:

1. We'll begin by obtaining the molar mass of CH4. This is illustrated below:

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

Number of mole of CH4 from the question = 5 moles

Mass of CH4 =?

Mass = number of mole x molar Mass

Mass of CH4 = 5 x 16

Mass of CH4 = 80g

2. Mass of O2 from the question = 38g

Molar Mass of O2 = 16x2 = 32g/mol

Number of mole O2 =?

Number of mole = Mass /Molar Mass

Number of mole of O2 = 38/32

Number of mole of O2 = 1.188mole

6 0
3 years ago
Find the percent composition of OXYGEN in Manganese (III) nitrate, Mn(NO3)3.
BaLLatris [955]

Answer:

59.8%

Explanation:

First find the Mr of manganese (III) nitrate.

Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>

Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:

Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>

Now we can find percentage composition / percentage by mass of oxygen.

% composition = \frac{Mr\ of\ oxygen\ in\ compound}{Mr\ of\ compound} × 100

% composition = \frac{144}{240.9} × 100 = <u>59.776%</u>

∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).

8 0
2 years ago
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