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2 years ago

How much does a ram pickup cost asking for a friend

Chemistry

1 answer:

nika2105 [10]2 years ago

4 0

Answer:alot lol

Explanation:

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Which element would be a strong reducing agent? F2 Ba Na Cl2

Damm [24]

F2 because it’s the strongest

3 0

2 years ago

A gas with a volume of 500. mL at 75°C is heated to 225 °C. What is the new volume of the gas?

mihalych1998 [28]

Using v1/t1=v2/t2
v1=500
v2=?
t1=75=368k
t2=225=498
500/368=v2/498
1.4x498=v2
v2=697.2ml

6 0

2 years ago

I have my chemistry EOC tomorrow, and I need to know how to calculate the molar mass of a compound. Specifically ones set up lik

yuradex [85]

Answer:

357.475

Explanation:

First you need periodic table and you have to look for mass

Fe = 3 x 55.845 = 167.535

P = 2 x 30.97 = 61.94

o = 4 x 2 so 8 oxygen = 8 x 16 = 128

add all and you get 357.475

6 0

2 years ago

What is the molarity of 70.6 g C2H6O in 2.25 L of solution

myrzilka [38]

Hey there!

Molar mass C2H6O = 46.0684 g/mol

Number of moles:

n = mass of solute / molar mass

n = 70.6 / 46.0684

n = 1.532 moles

Therefore:

M = number of moles / volume ( L )

M = 1.532 / 2.25

= 0.680 M

Hope that helps!

7 0

3 years ago

How many moles of O2 are required to react with 6.6 moles of H2?

Bingel [31]

Answer:

For 1: 3.3 moles of oxygen gas is required.

For 2: 14 moles of hydrogen gas is required.

For 3: 1.5 moles of oxygen gas is required.

Explanation:

The chemical reaction of oxygen and hydrogen to form water follows:

$O_2+2H_2\rightarrow 2H_2O$

For 1: When 6.6 moles of $H_2$ is reacted.

By Stoichiometry of the above reaction:

2 moles of hydrogen gas reacts with 1 mole of oxygen gas.

So, 6.6 moles of hydrogen gas will react with = $\frac{1}{2}\times 6.6=3.3mol$ of oxygen gas.

Hence, 3.3 moles of oxygen gas is required.

For 2: When 7.0 moles of $O_2$ is reacted.

By Stoichiometry of the above reaction:

1 mole of oxygen gas reacts with 2 moles of hydrogen gas.

So, 7 moles of oxygen gas will react with = $\frac{2}{1}\times 7=14mol$ of hydrogen gas.

Hence, 14 moles of hydrogen gas is required.

For 3: When 3.0 moles of $H_2O$ is formed.

By Stoichiometry of the above reaction:

2 moles of water is formed from 1 mole of oxygen gas.

So, 3.0 moles of water will be formed from = $\frac{1}{2}\times 3.0=1.5mol$ of oxygen gas.

Hence, 1.5 moles of oxygen gas is required.

7 0

2 years ago