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Serhud [2]
3 years ago
10

The Half-Life of Pa-234 is 6.75 hours. How much (what fraction) of a sample of this isotope remains after 20.25 hours?

Chemistry
1 answer:
Grace [21]3 years ago
6 0
So in one hour half the amount remains (that's what half life means). In two hours 1/4 (or half of half) would remain and in three hours 1/8 would remain.
So the answer is 1/8
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zalisa [80]

Answer:

Radio waves. Is the answer

6 0
3 years ago
Hurry please!! Needs to be turned in by 20 minutes!! Describe the two main factors the influence the path of surface currents. W
Volgvan

Currents on the surface are determined by three major factors: the major overall global wind patterns, the rotation of the Earth, and the shape of ocean basins.

7 0
3 years ago
Number of grams of hydrogen than can be prepared from 6.80g of aluminum​
Anastaziya [24]

Answer:

0.7561 g.

Explanation:

  • The hydrogen than can be prepared from Al according to the balanced equation:

<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>

It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.

  • Firstly, we need to calculate the no. of moles of (6.8 g) of Al:

no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.

<em>Using cross multiplication:</em>

2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.

0.252 mol of Al need to react → ??? mol of H₂.

∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.

  • Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:

mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.

5 0
4 years ago
3 H2SO4(aq) + 2 Al(s) → Al2(SO4)3(aq) + 3 H2(g) How many moles of Al2(SO4)3 are produced when 6.5 moles of aluminum are consumed
Lera25 [3.4K]

Answer:

520 moles were produced

7 0
3 years ago
What temperature kelvin would .97mol of gas be to occupy 26L at 751.2mmHg?
crimeas [40]

Answer: 322.56 Kelvin

Explanation:

Use the Ideal Gas Law

PV=nRT

R is the gas constant

T is the temperature in Kelvins

P is the pressure in atmospheres

V is the volume in liters

n is the number of moles of gas

First, the mm of mercury need to be converted to atmospheres using the conversion factor 1atm = 760 torr.

751.2mmHg (torr) =0.988atm

Now plug everything in

(0.988)(26)=(0.97)(0.0821)T\\T=322.56K

5 0
3 years ago
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