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kodGreya [7K]
3 years ago
9

A 100g sample of Carbon-14 has a half-life of 5 years. How much Carbon-14 is left after 10 years?

Chemistry
1 answer:
borishaifa [10]3 years ago
5 0

Answer:  The carbon-14 left after 10 years is 25 g

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) for rate constant

t_{\frac{1}{2}}=\frac{0.693}{k}

k=\frac{0.693}{5years}=0.139years^{-1}

b) for amount left after 10 years

t=\frac{2.303}{k}\log\frac{a}{(a-x)}

10=\frac{2.303}{0.139}\log\frac{100}{(a-x)}

(a-x)=25g

Thus carbon-14 left after 10 years is 25 g

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Define shielding and effective nuclear charge. What is the connection between the two?
ivolga24 [154]

Answer:

The relation between the shielding and effective nuclear charge is given as

Z_{eff} = Z -S

where s denote shielding

z_{eff} denote effective nuclear charge

Z - atomic number

Explanation:

shielding is referred to as the repulsion of an outermost electron to the pull of electron from valence shell.  Higher the electron in valence shell higher will be the shielding effects.  

Effective nuclear charge is the amount of net positive charge that valence electron has.

The relation between the shielding and the effective nuclear charge is given as  

Z_{eff} = Z -S

wheres denote shielding

z_{eff} denote effective nuclear charge  

Z - atomic number

8 0
3 years ago
How many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of Al and 1
valentinak56 [21]
<span>4 Al + 3 O2 → 2 Al2O3 

(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al 
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2 

0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess. 

((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) = 
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7 0
3 years ago
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At equilibrium at 2500K, [HCl]=0.0625M and [H2]=[Cl2]=0.00450M for the reaction H2+Cl2 ⇌ HCl.
ASHA 777 [7]

Answer:

a. H_2+Cl_2 \rightleftharpoons 2HCl

b. K = 192.9

c. Products are favored.

Explanation:

Hello!

a. In this case, according to the unbalanced chemical reaction we need to balance HCl as shown below:

H_2+Cl_2 \rightleftharpoons 2HCl

In order to reach 2 hydrogen and chlorine atoms at both sides.

b. Here, given the concentrations at equilibrium and the following equilibrium expression, we have:

K=\frac{[HCl]^2}{[H_2][Cl_2]}

Therefore, we plug in the data to obtain:

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c. Finally, we infer that since K>>1 the forward reaction towards products is favored.

Best regards!

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2 years ago
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4 0
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Answer:

1

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we take tailing zeroes as significant figures only in case of decimals

3 0
3 years ago
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