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3 years ago

If you start with 100 grams of hydrogen-3, how many grams will you have after 24.6 years?

Chemistry

1 answer:

svet-max [94.6K]3 years ago

4 0

Answer:

The mass left after 24.6 years is 25.0563 grams

Explanation:

The given parameters are;

The mass of the hydrogen-3 = 100 grams

The half life of hydrogen-3 which is also known as = 12.32 years

The formula for calculating half-life is given as follows;

$N(t) = N_0 \times \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{\frac{1}{2} }} }$

Where;

N(t) = The mass left after t years

N₀ = The initial mass of the hydrogen-3 = 100 g

t = Time duration of the decay = 24.6 years

$t_{\frac{1}{2} }$ = Half-life = 12.32 years

$N(24.6) = 100 \times \left (\dfrac{1}{2} \right )^{\dfrac{24.6}{12.32}} } = 25.0563$

The mass left after 24.6 years = 25.0563 grams.

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Why does increasing the temperature make a solid dissolve faster?

Deffense [45]

Answer:

The chemical bonds of the solid are broken faster.

Explanation:

As the temperature of a solution is increased, the average kinetic energy of the molecules that make up the solution also increases. The increased vibration (kinetic energy) of the solute molecules causes them to be less able to hold together, and thus they dissolve more readily.

6 0

2 years ago

A 11.0 mLmL sample of 0.30 MHBrMHBr solution is titrated with 0.16 MNaOHMNaOH. Part A What volume of NaOHNaOH is required to rea

cupoosta [38]

Answer:

21 mL of NaOH is required.

Explanation:

Balanced reaction: $HBr+NaOH\rightarrow NaBr+H_{2}O$

Number of moles of HBr in 11.0 mL of 0.30 M HBr solution

= $(\frac{0.30}{1000}\times 11.0)$ moles = 0.0033 moles

Let's say V mL of 0.16 M NaOH solution is required to reach equivalence point.

So, number of moles of NaOH in V mL of 0.16 M NaOH solution

= $(\frac{0.16}{1000}\times V)$ moles = 0.00016V moles

According to balanced equation-

1 mol of HBr is neutralized by 1 mol of NaOH

So, 0.0033 moles of HBr are neutralized by 0.0033 moles of NaOH

Hence, $0.00016V=0.0033$

$\Rightarrow V=\frac{0.0033}{0.00016}=21$

So, 21 mL of NaOH is required.

4 0

3 years ago

Helppppppp I have a brain fart ....

kherson [118]

Answer:

Hello there!

The first answer is hot

The second answer is sunshine!

Explanation:

Those make the most sense

Have a great day!

6 0

3 years ago

50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH sol

boyakko [2]

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid]) (Eq. 01)

Where

pKa = -log(Ka) (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2 (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get

pH = 3.35

5 0

3 years ago

An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its

german

Hello!

First you need to calculate q
delta U is change in internal energy 

delta U = q + w 
q is heat and w work done 
here work was done by the system means energy leaving the system so w is negative 

delta U = q + w 

q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ 

q = m x c x delta T 

7211 J = 80.0 g x c x (225-25) °C 

c = 0.451 J /g °C

Hope this Helps! Have A Wonderful Day! :)

5 0

3 years ago