1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Margarita [4]
3 years ago
12

FREE POINTS DONT FORGET TO GIVE A HEART ON MY ACCOUNT HELPS

Chemistry
2 answers:
fredd [130]3 years ago
8 0

Answer:

:) this is pretty cool my guy

Tju [1.3M]3 years ago
7 0

Answer:

pfffft-

Explanation:

Anime Pros like me, don't give hearts out, get real

You might be interested in
I WILL GIVE BRAINLIEST
Sergeeva-Olga [200]

Answer:

waste gas he should use algae

4 0
3 years ago
When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H
algol [13]

Answer: The empirical formula is CH_2 and molecular formula is C_4H_8

Explanation:

We are given:

Mass of CO_2 = 18.95 g

Mass of H_2O= 7.759 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 18.59 g of carbon dioxide, =\frac{12}{44}\times 18.59=5.07g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, =\frac{2}{18}\times 7.759=0.862g of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.422}{0.422}=1

For H =\frac{0.862}{0.422}=2

The ratio of C : H = 1: 2

Hence the empirical formula is CH_2.

The empirical weight of CH_2 = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4

The molecular formula will be=4\times CH_2=C_4H_8

6 0
3 years ago
How many moles of oxygen are present in 16 g of oxygen gas​
makkiz [27]
Hope this helps

Answer- 1 mole
4 0
3 years ago
Read 2 more answers
1. A gas sample at a pressure of 5.00 atm has a volume of 3.00 L. If the gas pressure is changed to 760 mm Hg, what volume will
tatyana61 [14]

Answer:

V₂ =  15.00 atm

Explanation:

Given data:

Initial pressure = 5.00 atm

Initial volume = 3.00 L

Final pressure = 760 mmHg ( 760/760 = 1 atm)

Final volume = ?

Solution:

P₁V₁ = P₂V₂

V₂ = P₁V₁ /  P₂

V₂ =  5.00 atm × 3.00 L / 1 atm

V₂ =  15.00 atm

8 0
3 years ago
What is a mole?
fredd [130]

Answer:

B.

Explanation:

One mole is the amount of substance that contain the Avogadro number which is equal to 6.022×10^23 atom, molecules or ions.

7 0
2 years ago
Other questions:
  • What were the first galaxies made of
    8·1 answer
  • In 1687, ___ explained that gravitational forces hold the solar system together.
    15·2 answers
  • What causes water to become denser when it is carried to the poles by surface currents? increased temperature and increased sali
    12·2 answers
  • 1 calorie is equal to<br><br> A. 4.184 J<br> B. 1,000 J<br> C. 6.02x10^23 J<br> D. 8.314 J
    11·1 answer
  • The addition of NaOH to water would result in which of the following?
    11·1 answer
  • How do hydrogen bonds make water a liquid?
    7·1 answer
  • 1. Identify all of the gas law equations that relate to the ideal gas law.
    6·1 answer
  • Please help im being timed :((
    6·1 answer
  • if a saline bag has a molarity of 0.155 and 0.75 L are given to a patient how many grams of NaCl did the patient receive?
    9·1 answer
  • Compare the properties of elements in groups 17 and group 18. Make sure to cover the following points:
    7·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!