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slamgirl [31]
3 years ago
7

20 POINTS 2 QUESTIONS CAN YOU ANSWER THEM ???

Chemistry
2 answers:
Viktor [21]3 years ago
7 0
2. <span>1/8 = (1/2)³ </span>
<span>so it's 3 half lives.</span>
Reika [66]3 years ago
4 0

<u>Answer 1 :</u> The synthesis reaction is, N_2+3H_2\rightarrow 2NH_3

Explanation :

Synthesis reaction : It is a type of reaction in which the reactants present in elemental state react to give a single product.

(1) N_2+3H_2\rightarrow 2NH_3

This is a synthesis reaction.

(2) MgI_2+Br_2\rightarrow MgBr_2+I2

This is a single displacement reaction in which the most reactive element displaces the least reactive element.

(3) KClO_3\rightarrow KCl+O_2

This is a decomposition reaction in which larger reactant decomposes to give two or more products.

(4) HCl+NaOH\rightarrow NaCl+H_2O

This is a neutralization reaction in which an acid and a base react to give a salt and water as a product.

<u>Answer 2 : </u>The number of half-lives it take will be, 3

Formula used :

a=\frac{a_o}{2^n}

where,

a = amount of reactant left after n-half lives = \frac{1}{8}a_o

a_o = Initial amount of the reactant

n = number of half lives

Putting values in above equation, we get:

\frac{1}{8}a_o=\frac{a_o}{2^n}

2^n=8=2^3

n = 3

Therefore, the number of half-lives it take will be, 3

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Which of these are ionic compounds? Check all that apply.
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What is the ultimate end of a high mass star that has a core about 2.8x the mass of our sun?
jolli1 [7]

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A

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3 years ago
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aseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.5 g of methane is
maksim [4K]

Answer:

There is 9.6 grams of CO2 produced

Explanation:

Step 1: Data given

Mass of methane = 5.50 grams

Molar mass of methane = 16.04 g/mol

Mass of oxygen = 13.9 grams

Molar mass of oxygen = 32.0 g/mol

Step 2: The reaction

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles methane = 5.50 grams / 16.04 g/mol

Moles methane = 0.343 moles

Moles oxygen = 13.9 grams / 32.0 g/mol

Moles oxygen = 0.434 moles

For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O

O2 is the limiting reactant. It will completely react (0.434 moles).

There will react 0.434/2 = 0.217 moles CH4

There will remain 0.343-0.217 = 0.126 moles CH4

There will be produced 0.434 moles of H2O and

0.434/2 =0.217 moles of CO2

Step 4: Calculate mass of products

Mass = moles * molar mass

Mass CO2 = 0.217 moles ¨44.01 g/mol

Mass CO2 = 9.6 grams

Mass H2O = 0.434 moles * 18.02

Mass H2O = 7.8 grams

4 0
3 years ago
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