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andreyandreev [35.5K]
3 years ago
8

Help it’s due at a certain time

Chemistry
1 answer:
DiKsa [7]3 years ago
3 0

Answer:

b

Explanation:

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What is helium an example of? <br> element <br> compound <br>atomic number <br>subatomic particle​
fgiga [73]

Answer:

atomic number

Explanation:

hope this helps

4 0
3 years ago
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A piece of wood with a density of 0.85 g/cm^3 has a mass of 25 g. what is the volume of the wood?
Karolina [17]
Density = m/v  therefore    v = m/d     v = 25/0.85    v = 29.4117... cm^3 ( = 2.94 * 10^-5 m^3
5 0
3 years ago
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How do you solve this? I am so confused. This is Chemistry 1 HON and its dimensional analysis. It would help me so much if someo
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4 0
3 years ago
A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO
Akimi4 [234]

Answer: The empirical formula for the given compound is CH_2

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of CO_2=22.0g

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, \frac{12}{44}\times 22.0=6g of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.5 moles.

For Carbon = \frac{0.5}{0.5}=1

For Hydrogen  = \frac{1.0}{0.5}=2

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is C_{1}H_{2}=CH_2

4 0
3 years ago
I am having some trouble figuring out how to approach the following problem: A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is ti
Tanya [424]
<span>We look at the end of the day:

n(HNO3) added = 0.500*17.0/1000 = 0.00850 mol
n(NH3) = 0.200*75.0/1000 - 0.00850 = 0.00650 mol
[NH3] left = 0.00650*1000/(17.0+75.0) = 0.070652
M [OH-] = Kb * [NH3] = 0.070652*1.8*10^(-5) = 1.27174 x 10^(-6)
pOH = -log[OH-] ≈ 5.8956 pH = 14 - pOH ≈ 8.10</span>
4 0
3 years ago
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