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svetlana [45]
3 years ago
8

A neutral atom posses an atomic number of 13 and an atomic mass of 27 three electrons are lost. From which regio of an atom arr

they lost
Chemistry
1 answer:
AleksAgata [21]3 years ago
8 0
Theyre lost from the atomic number
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Which of the following is similar among all living organisms on Earth?
statuscvo [17]

A. cellular function

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5 0
3 years ago
The photodissociation of ozone by ultraviolet light in the upper atmosphere is a first-order reaction with a rate constant of 1.
atroni [7]

Answer:

[O₃]= 8.84x10⁻⁷M  

Explanation:

<u>The photodissociation of ozone by UV light is given by:</u>

O₃ + hν → O₂ + O (1)

<u>The first-order reaction of the equation (1) is:</u>

rate = k [O_{3}] = - k \frac{\Delta [O_{3}]}{\Delta t} (2)

<em>where k: is the rate constant and Δ[O₃]/Δt: is the variation in the ozone concentration with time, and the negative sign is by the decrease in the reactant concentration </em>    

<u>We can get the following expression of the </u><u>first-order integrated law</u><u> of the reaction (1), by resolving the equation (2):</u>

[O_{3}]_{t} = [O_{3}]_{0} \cdot e^{-kt} (3)

<em>where [O₃](t): is the ozone concentration in the elapsed time and [O₃]₀: is the initial ozone concentration</em>

We can calculate the initial ozone concentration using equation (3):  

[O_{3}]_{t} = 5.0 \cdot 10^{-3}M \cdot e^{-(1.0\cdot 10^{-5}s^{-1}) (\frac{10d \cdot 24h \cdot 3600 s}{1d \cdot 1h})} = 8.84 \cdot 10^{-7}M

So, the ozone concentration after 10 days is 8.84x10⁻⁷M.

I hope it helps you!                    

3 0
3 years ago
The percent by mass of copper in CuBr2 is
Sunny_sXe [5.5K]
Hope this helps you.

4 0
2 years ago
Read 2 more answers
Phosphorous pentoxide, P2O5(s), is produced from the reaction between pure oxygen and pure phosphorous (P, solid). What is the v
BartSMP [9]

Answer:

4190.22 L = 4.19 m³.

Explanation:

  • For the balanced reaction:

<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>

It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>

  • Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:

no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.

  • Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:

<u><em>Using cross multiplication:</em></u>

5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.

??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.

∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.

  • Finally, we can get the volume of oxygen using the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 60.95 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).

∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.

3 0
3 years ago
A metal X from two oxide A and B .3.oogm of A and B contain 0.72 and 1.16g of oxygen respectively.calculate the maases of metal
devlian [24]

Answer:

Explanation:

Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.

Metal X can form 2 oxides (A and B).

A + B = 3g

The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.

The mass of metal X in the two oxides will be the same because it's the same metal.

Thus, we represent the mass of the metal in the two oxides as 2X.

2X + 0.72 + 1.16 = 3

2X + 1.88 = 3

2X = 3 - 1.88

2X = 1.12

X = 0.56

<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>

Thus, mass of metal (X) in 1g of oxygen in A is

0.56g ⇒ 0.72g

X ⇒ 1

X = 1 × 0.56/0.72

X = 0.78 g

Hence, 0.78g of the metal will combine with 1g of oxygen for A

Also, mass of metal (X) in 1g of oxygen in B is

0.56g ⇒ 1.16g

X ⇒ 1g

X = 1×0.56/1.16

X = 0.48 g

Thus, 0.48g of the metal will combine with 1g of oxygen for B

6 0
2 years ago
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