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Veronika [31]
3 years ago
8

Could someone please help me with my chemistry hw please please please please please please please

Chemistry
1 answer:
IceJOKER [234]3 years ago
3 0
Jhgpqurhubfjnqjejfnbouqejdnfcnadvq f/g
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What does scientist use to deign their experiments
joja [24]

Answer:

a chart

Explanation:

a chart a chart is the answer

8 0
2 years ago
Hydrogen peroxide (h​2​o​2​) decomposes according to the equation:h​2​o​2​(​l​) ​⇆​ h​2​o(​l​) + ½ o​2​(​g​)calculate k​p​ for t
goldenfox [79]
Hello! Let me try to answer this :)

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6 0
3 years ago
When CO2(g) is put in a sealed container at 730 K and a pressure of 10.0 atm and is heated to 1420 K , the pressure rises to 24.
d1i1m1o1n [39]

Answer:

48%

Explanation:

Based on Gay-Lussac's law, the pressure is directly proportional to the temperature. To solve this question we must assume the temperature increases and all CO2 remains without reaction. The equation is:

P1T2 = P2T1

<em>Where Pis pressure and T absolute temperature of 1, initial state and 2, final state of the gas:</em>

P1 = 10.0atm

T2 = 1420K

P2 = ?

T1 = 730K

P2 = 10.0atm*1420K / 730K

P2 = 19.45 atm

The CO2 reacts as follows:

2CO2 → 2CO+ O2

Where 2 moles of gas react producing 3 moles of gas

Assuming the 100% of CO2 react, the pressure will be:

19.45atm * (3mol / 2mol) = 29.175atm

As the pressure rises just to 24.1atm the moles that react are:

24.1atm * (2mol / 19.45atm) = 2.48 moles of gas are present

The increase in moles is of 0.48 moles, a 100% express an increase of 1mol. The mole percent that descomposes is:

0.48mol / 1mol * 100 = 48%

8 0
3 years ago
How do I solve this problem?
JulijaS [17]
First question. Applying ideal gas equation PV=nRT, P= 101.3 x 10³Pa = 1atm. therefore, 1 x 260 x 10^-3 = n x 0.082 x 294.( Temperature in kelvin=273+21). n = 0.01 moles. Volume of gas at STP= n x 22.4 = 0.01x22.4 = 0.224L. Hope this helps
5 0
2 years ago
The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?​
Natasha_Volkova [10]

Answer:

3.13%.

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 0.15

Half life (t½) = 2 mins

Number of half-life (n) = 5

Fraction of sample remaining =.?

Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:

Amount remaining (N) = 1/2ⁿ × original amount (N₀)

NOTE: n is the number of half-life.

N = 1/2ⁿ × N₀

N = 1/2⁵ × 0.15

N = 1/32 × 0.15

N = 0.15/32

N = 4.69×10¯³

Therefore, 4.69×10¯³ is remaining after 5 half-life.

Finally, we shall the fraction of the sample remaining after 5 half-life as follow:

Original amount (N₀) = 0.15

Amount remaining (N) = 4.69×10¯³

Fraction remaining = N/N₀ × 100

Fraction remaining = 4.69×10¯³/0.15 × 100

Fraction remaining = 3.13%

3 0
3 years ago
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