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2 years ago

Generally, does the size of a radioactive sample affect half-life?

2 answers:

8 0

The quantity of radioactive nuclei at any given time will decrease to half as much in one half-life.

It is impossible to predict when an individual radioactive atom will decay

Natasha2012 [34]2 years ago

4 0

Answer:

The quantity of radioactive nuclei at any given time will decrease to half as much in one half-life.

Explanation:

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How many grams of helium must be released to reduce the pressure to 65 atmatm assuming ideal gas behavior? Express the answer in

lyudmila [28]

The question is incomplete, here is the complete question:

How many grams of helium must be released to reduce the pressure to 65 atm assuming ideal gas behavior. Note : 334-mL cylinder for use in chemistry lectures contains 5.209 g of helium at 23°C.

Answer: The mass of helium released is 1.6 grams

Explanation:

We are given:

Mass of helium in the cylinder = 5.209 g

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{w}{M}RT$

where,

P = Pressure of the gas = 65 atm

V = Volume of the gas = 334 mL = 0.334 L (Conversion factor: 1 L = 1000 mL)

w = Weight of the gas = ?

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gas = $23^oC=[23+273]K=296K$

Putting values in above equation, we get:

$65atm\times 0.334L=\frac{w}{4g/mol}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 296K\\\\w=\frac{65\times 0.334\times 4}{0.0821\times 296}=3.573g$

Mass of helium released = (5.209 - 3.573) g = 1.636 g = 1.6 g

Hence, the mass of helium released is 1.6 grams

5 0

3 years ago

What is nourishing for a seed of a plant?

fiasKO [112]

making sure that it has all the necessary conditions to become a big plant:

enough water
sunlight
good replenished soil

hope it helps...!!!

3 0

1 year ago

Read 2 more answers

Let’s be friends jajajajajajajajjajajajajajaja

Nutka1998 [239]

Ok bestyyyy :))))).............

4 0

3 years ago

Coal gasification can be represented by the equation: 2 C(s) + 2 H2O(g) → CH4(g) + CO2(g) ΔH = ? Use the following information t

svlad2 [7]

Answer: The $\Delta H^o_{rxn}$ for the reaction is 15 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$2C(s)+2H_2O(g)\rightarrow CH_4(g)+CO_2(g)$ $\Delta H^o_{rxn}=?$