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dimulka [17.4K]
3 years ago
12

Hi does anyone know how to do chem cause I might need a tut.or or sum.

Chemistry
2 answers:
Alenkinab [10]3 years ago
8 0

Answer:

I wish I can help, but I don't know chem. I know a website where there is always someone to help you and asking a question is free. You can go to Wyzant.com

Explanation:

Hope you do well on your Chem work. :)

scoray [572]3 years ago
6 0

Ask your parents or guardian for a tutor.

It's very dangerous to give someone your snap online that you don't know!

Have a nice day <3

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Calculate the number of milligrams to 0.425 kg
sp2606 [1]

Answer:

Explanation:

as we know that 1kg =1,000,000 mg

therefore

0.425 kg= 0.425*1,000,000/1=425000

result:

0.425 kg contain 425000 mg

i hope this will help you

4 0
3 years ago
A quantity of N2 gas originally held at 5.23 atm pressure in a 1.20 −L container at 26 ∘C is transferred to a 14.5 −L container
lara31 [8.8K]

Answer:

n (N₂) =  0.256 mol

n (O₂) = 1.0848 mol

n (Total) = 1.3408 mol

Pressure in new Container =  2.222 atm

Explanation:

Data Given:

For Nitrogen gas (N₂)

Pressure of N₂ gas =  5.23 atm

Volume of N₂ gas = 1.20 L

Temperature of N₂ gas = 26° C

Temperature of N₂ gas in Kelven (K) = 26° C +273

Temperature of N₂ gas in Kelven (K) = 299K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

quantity of gas N₂ gas = ?

For Oxygen gas (O₂)

Pressure of O₂ gas =  5.21 atm

Volume of O₂ gas = 5.21 L

Temperature of O₂ gas = 26° C

Temperature of O₂ gas in Kelven (K) = 26° C +273

Temperature of O₂ gas in Kelven (K) = 299K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

quantity of gas O₂ gas = ?

*we also have to find the total Pressure in the new container = ?

Formula Used

                         PV =nRT

                        n (N₂) = PV /RT . . . . . . . . . . . . . (1)

* Find the quantity of N₂

Put value in formula (1)

                n (N₂) = 5.23 atm x 1.20 L / 0.08206 L atm K⁻¹ mol⁻¹ x 299K

                n (N₂) =  6.276 atm .L /  24.52 L atm. mol⁻¹ x 299K

                n (N₂) =  0.256 mol

* Find the quantity of O₂

Put value in formula (1)

                n (O₂) = 5.21 atm x 5.10 L / 0.08206 L atm K⁻¹ mol⁻¹ x 299K

                n (O₂) =  26.6 atm .L /  24.52 L atm. mol⁻¹

                n (O₂) = 1.0848 mol

*Now to find the Total Quantity of both gases

                n(Total) =  n (N₂) + n (O₂)

                 n (Total) = 0.256 mol + 1.0848 mol

                 n (Total) = 1.3408 mol

**To find the Total Pressure in the new Container

Data to calculate Total Pressure in new container

Volume of gas = 14.5 L

Temperature of gases = 20° C

Temperature of gases in Kelven (K) = 20° C +273

Temperature of gases in Kelven (K) = 293K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

Volume Pressure in new container = ?

Formula Used

                         PV =nRT

                        P = nRT / V . . . . . . . . . . . . . (2)

Put values in Equation (2)

           P =  1.3408 mol x 0.08206 L atm K⁻¹ mol⁻¹ x 293 K / 14.5 L

           P =  2.222 atm

8 0
4 years ago
How much of a 60.0 gram sample of Carbon-14 DECAYED after 17,100 if 5
zlopas [31]
The answer is A.7.50
3 0
3 years ago
Complete the following reaction: Li + S --&gt;
valina [46]

Answer:

Li2S

Explanation:

Hope this helps.

7 0
3 years ago
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porque a água da chuva, mesmo em ambientes não poluídos, pode ser considerada levemente ácida?indique a fórmula do composto quím
Basile [38]

Answer:

Olá, a resposta que você está procurando é, na verdade, plasma mais refração. Eu estou usando um tradutor para ajudá-lo agora, então eu espero que você não se importe.

Explanation:

5 0
4 years ago
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