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RUDIKE [14]
3 years ago
7

The density of helium in a balloon is 1.18 g/L. If a

Chemistry
1 answer:
marissa [1.9K]3 years ago
3 0
There is 3.58 He in the balloon.
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Fe(NO3)2 not sure how to get the oxidation numbers of all elements
Cloud [144]
<span>Fe(NO3)2
The NO3 part is a poly-atomic ion with total charge -1.
This is because Fe has a +2 charge and two NO3's with a -1 charge will balance out to 0.
 Most often we just make the assumption that Oxygen has a -2 oxidation number because it is very electro-negative.
So to find N, we just need an oxidation number that balances out with 3(-2) to get -1 (the total charge of the ion)</span>
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3 years ago
Calculate the mass of 4.60 x 10^25 atoms of neon?
enyata [817]
I think it’s B 5.54 x 10^2g
4 0
3 years ago
A 1.20-L container contains 1.10 g of an unknown gas at STP. What is the molecular weight of the unknown gas?
SOVA2 [1]

Answer:

M = 20.5 g/mol

Explanation:

Given data:

Volume of gas = 1.20 L

Mass of gas = 1.10 g

Temperature and pressure = standard

Solution:

First of all we will calculate the density.

Formula:

d = mass/ volume

d = 1.10 g/ 1.20 L

d = 0.92 g/L

Now we will calculate the molar  mass.

d = PM/RT

0.92 g/L = 1 atm × M / 0.0821 atm.L/mol.K ×273.15 K

M =  0.92 g/L × 0.0821 atm.L/mol.K ×273.15 K /  1 atm

M = 20.5 g/mol

8 0
3 years ago
Mole are in 5.1 grams of beryllium ?
chubhunter [2.5K]
5.1/9.01 ≈ 0.566 moles
7 0
3 years ago
How much heat is lost when 1277 g of H20 at 400 K is changed into ice at 263 K? Plz help I’ll give you brainliest 20 points
OLga [1]

The heat lost is -7.32*10^-^5J

The heat lost when the ice is cooled from 400k to 263K can be calculated using the formula of heat transfer.

<h3>Heat Transfer</h3>

This is the heat transferred from a body of higher temperature to a body of lower temperature.

Q = mc(\delta)T

  • Q = Heat Transfer
  • m = mass = 1277g
  • ΔT  = change in temperature

\delta T = (400 - 273) - (263 - 273) = \\&#10;\delta T = T_2 - T_1\\&#10;\delta T = -10 - 127\\&#10;\delta T = -137^0C

We converted the temperature from kelvin scale into Celsius scale and find the change in temperature.

Solving for heat transfer

Q = mc\delta T\\&#10;Q = 1277 * 4.186 * -137\\&#10;Q = -732336.514J

The heat loss is approximately -7.32*10^-^5J

Learn more on heat transfer here;

brainly.com/question/16055406

6 0
2 years ago
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