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IrinaK [193]
3 years ago
5

When an atom loses one or more electrons, this atom becomes a

Chemistry
1 answer:
n200080 [17]3 years ago
5 0
It becomes a positive ion
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Assessment
MakcuM [25]

i didn’t mean to put this opps

4 0
3 years ago
Two gases X and Y are found in the atmosphere in only trace amounts because they decompose quickly. When exposed to ultraviolet
igomit [66]

Answer:

1. Yes

2. After 68.1 mins, pX < pY.

Explanation:

Assuming the total gas pressure is 1 atm, let the partial pressure of Y be y, partial pressure of X will be 0.25y + y = 1.25y

1.25y + y = 1atm

2.25y = 1 atm

y = 1 atm / 2.25 = 0.44 atm

Then partial pressure of X = 0.56 atm

The partial pressure of a gas in a mixture of gases is directly proportional to the mole fraction of the gas.

Therefore the mole fraction of X and Y is 0.56 and 0.44 respectively.

The partial pressures of X and Y becomes half of their original values at 1.25 h = 85 min and Y at 150 min respectively.

The partial pressure after some time can be calculated from the half-life equation :

m = m⁰ *  1/2ⁿ

Where m = the remaining mass, m⁰ = initial mass, and n is number of half-lives undergone.

Let partial pressures represent the mass, and n for X and Y be a and b respectively:

pX  = 0.56/2ᵃ

pY = 0.44/2ᵇ

We then determine when Partial pressure of X, pX = Partial pressure of Y, pY

0.56/2ᵃ = 0.44/2ᵇ

2ᵃ/2ᵇ = 0.56/0.44

2ᵃ/2ᵇ = 1.27

2ᵃ⁻ᵇ = 2⁰°³⁴⁵

a - b = 0.345

Let this time be t, therefore,

For X; t = 85a and For Y: t = 150b

85a = 150b

then, a = 1.76b

1.76b - b = 0.345

0.760b = 0.345

b = 0.454 and,

a = 0.345 + 0.454 = 0.799

So,  X goes through 0.779 half-lives while Y goes through 0.454 half-lives Then, the time for both X and Y to have the same amount is:

t = 150 * 0.454 = 68.1 min

After 68.1 mins, pX < pY.

5 0
3 years ago
A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →
vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

7 0
3 years ago
Identify the element that cannot participate in nuclear fission reactions. (Hint: Think about the size of the atom.)
chubhunter [2.5K]
A or C I’m pretty sure
8 0
3 years ago
Which statement correctly describes diamond and graphite, which are different forms of solid carbon?
mart [117]
They differ in their molecular structures and properties.
5 0
3 years ago
Read 2 more answers
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