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ZanzabumX [31]
3 years ago
11

A magnet is an example of a

Chemistry
1 answer:
statuscvo [17]3 years ago
7 0
The answer is distance force
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Complete combustion of 7.50 g of a hydrocarbon produced 23.1 g of CO2 and 10.6 g of H2O. What is the empirical formula for the h
r-ruslan [8.4K]
A CH compound is combusted to produce CO2 and H2O 
CnHm + O2 -----> CO2 + H2O 
Mass of CO2 = 23.1g
 Mass of H2O = 10.6g
 Calculate by mass of the compounds
 For Carbon C, divide by molecular weight of CO2 and multiply with Carbon
molecular weight. So C in grams = 23.1 x (12.01 / 44.01) = 6.3 g C
 For Hydrogen H, divide by molecular weight of H2O and multiply with Hydrogen molecular weight. So H in grams = 10.6 x (2.01 / 18.01) = 0.53 g C
= 1.18 of H
 Calculate the moles for C and H
 6.3 grams of C x (1 mole/12.01 g C) = 0.524 moles of C
 1.18 grams of H x (1 mole/1.008 g H) = 1.17 moles of H
 Divides by both mole entities with smallest
 C = 0.524 / 0.524 = 1 x 4 = 4
 H = 1.17 / 0.524 = 2.23 x 4 = 10
 The empirical formula is C4H10.
7 0
3 years ago
Read 2 more answers
4. Which of the following NOT TRUE about real gas
Semenov [28]

Answer:

The answer is D.

Explanation:

Intermolecular force are negligible

When the distance between molecules decrease,

the attraction or repulsion become greater

7 0
2 years ago
P4 + O₂<br> .P₂ Oz<br> what is the balance
34kurt
I am not sure sorry
7 0
2 years ago
You want to determine the concentration of a solid solute dissolved in a specified volume of liquid solvent which of the followi
Natali [406]
Generally, chemists prefer to use morality (B) because it only invovles measuring the final volume of the solution and amount of moles of the solute

Hope this helps
6 0
3 years ago
Read 2 more answers
15.00 g of NH4HS(s) is introduced into a 500. mL flask at 25 °C, the flask is sealed, and the system is allowed to reach equilib
Ahat [919]

Answer:

0.328 atm

Explanation:

Kp is the equilibrium constant calculated based on the pressure, and it depends only on the gas substances. It will be the multiplication of partial pressures of the products raised to their coefficients divided by the multiplication of partial pressures of the reactants raised to their coefficients.

For the equation given, the stoichiometry is 1 mol of NH₃ for 1 mol of H₂S, so they will have the same partial pressure in equilibrium, let's call it p. So:

Kp = pxp

0.108 = p²

p = √0.108

p = 0.328 atm, which is the partial pressure of the ammonia.

3 0
3 years ago
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