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3 years ago

15

The half-life of radium-226 is 1,600 years. It decays into radon-222. What fraction of the original amount of radium-226 in a sa

mple will still be radium
after 8,000 years?
A.1/8
B.1/4
c.1/32
D.1/16

2 answers:

Mnenie [13.5K]3 years ago

6 0

Answer:

Your answer will be C. 1/32

vfiekz [6]3 years ago

5 0

Answer:

C. 1/32

Explanation:

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How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --> __ AlCl3 + __

kiruha [24]

Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

determine the number of moles of HCl
determine the mole ratio,
use the mole ratio to calculate the number of moles of aluminum.
use RFM of Aluminum to determine the grams required.

<u>Moles </u><u>of </u><u>HCl</u>

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

$x \: mol \: = \: \frac{2 \: \times \: 35}{1000} \\ = 0.07 \: moles \:$

We have 0.07 moles of HCl.

<u>Mole </u><u>ratio</u>

Balanced equation

6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)

Hence mole ratio = 6 : 2 (HCl : Al

but moles of HCl is 0.07, therefore the moles of Al;

$= \frac{2}{6} \times 0.07 \\ \: = 0.0233333 \: moles$

Therefore we have 0.0233333 moles of aluminum.

<u>Grams of </u><u>Aluminum</u>

We use the formula;

$grams \: = moles \: \times \: rfm$

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

$= 0.0233333 \: moles \: \times \: 26.982 \: \frac{g}{mol} \\ = 0.625799 \: grams$

The number of grams of aluminum required to react with HCl is 0.6258 g.

3 0

2 years ago

What is the mass percentage of silicon?

Mkey [24]

Answer: 46.743%

Explanation: what is the mass percentage of silicon? 46.743%

***If you found my answer helpful, please give me the brainliest, please give a nice rating, and the thanks ( heart icon :) ***

8 0

3 years ago

Dos vehículos colisionan. Se sabe que antes del choque el vehículo 1 de masa "m", se desplazaba horizontalmente hacia la derecha

BabaBlast [244]

Answer:

translate? ill edit this later

Explanation:

8 0

3 years ago

Lawrencium-262 has a half-life of 4 hr. How much of a 40 mg sample remains after 12 hours?

CaHeK987 [17]

Answer:

5 mg

Explanation:

If one half life is 4 hours, then 3 half lives is 12 hours.

This means that the sample will decay to 1/8 of its original amount.

So, the answer is 40(1/8) = 5 mg.

3 0

2 years ago

Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying

PIT_PIT [208]

The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? ( $K_a$ for HF is $6.8\times 10^{-4}$ .)

[HF] = 0.280 M

Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = $K_a=6.8\times 10^{-4}$

$HF\rightleftharpoons H^++F^-$

Initially

c 0 0

At equilibrium :

(c-x) x x

The expression of disassociation constant is given as:

$K_a=\frac{[H^+][F^-]}{[HF]}$

$K_a=\frac{x\times x}{(c-x)}$

$6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}$

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

$[H^+]=x=0.01346 M$

The pH of the solution is ;

$pH=-\log[H^+]=-\log[0.01346 M]=1.87$

The pH of an 0.280 M HF solution is 1.87.

6 0

3 years ago