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3 years ago

Use lewis theory to determine the formula for the compound that forms from ca and s.

1 answer:

6 0

<span>The formula in terms of the oxidation numbers for the elements. The sum of the oxidation numbers for a compound must be zero. Ca3N2 KBr Cs2S CsF</span>

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An archaeologist graduate student found a leg bone of a large animal during the building of a new science building. The bone had

Vlad [161]

Answer : The time passed in years is $2.74\times 10^2\text{ years}$

Explanation :

Half-life of carbon-14 = 5730 years

First we have to calculate the rate constant, we use the formula :

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

$k=\frac{0.693}{5730\text{ years}}$

$k=1.21\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the time passed.

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 = $1.21\times 10^{-4}\text{ years}^{-1}$

t = time passed by the sample = ?

a = initial amount of the reactant disintegrate = 15.3

a - x = amount left after decay process = 14.8

Now put all the given values in above equation, we get

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{15.3}{14.8}$

$t=274.64\text{ years}=2.74\times 10^2\text{ years}$

Therefore, the time passed in years is $2.74\times 10^2\text{ years}$

4 0

3 years ago

name seven elements that have names and symbols that do not start with the same letter (include the symbol after the name)

viktelen [127]

Answer:

ok will do

Explanation:

5 0

2 years ago

Read 2 more answers

In an experiment, a student needs 250.0 mL of a 0.100 M copper (II) chloride solution. A stock solution of 2.00 M copper (II) ch

LekaFEV [45]

Answer : The volume of stock solution needed are, 12.5 mL

Explanation :

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the initial molarity and volume of copper (II) chloride.

$M_2\text{ and }V_2$ are the final molarity and volume of stock solution of copper (II) chloride.

We are given:

$M_1=0.100M\\V_1=250.0mL\\M_2=2.00M\\V_2=?$

Putting values in above equation, we get:

$0.100M\times 250.0mL=2.00M\times V_2\\\\V_2=12.5mL$

Hence, the volume of stock solution needed are, 12.5 mL

8 0

2 years ago

Can anyone please help with this!!

Sphinxa [80]

Answer:

2341, last option is the correct choice.

Explanation:

Boiling points of the given compounds are given as:

$CH_4=-161.5\:C^{\circ}\\\\C_4H_{10}=-1\:C^{\circ}\\\\CH_2Cl_2=39.6\:C^{\circ}\\\\H_2O=100\:C^{\circ}$

Best Regards!

5 0

3 years ago

Hc2h3o2(aq)+h2o(l)⇌h3o+(aq)+c2h3o−2(aq) kc=1.8×10−5 at 25∘c part a if a solution initially contains 0.260 m hc2h3o2, what is the

Cloud [144]

Answer : The equilibrium concentration of $H_3O^+$ at $25^oC$ is, $2.154\times 10^{-3}m$ .

Solution : Given,

Equilibrium constant, $K_c=1.8\times 10^{-5}$

Initial concentration of $HC_2H_3O_2$ = 0.260 m

Let, the 'x' mol/L of $H_3O^+$ are formed and at same time 'x' mol/L of $HC_2H_3O_2$ are also formed.

The equilibrium reaction is,

$HC_2H_3O_2(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+C_2H_3O_2^-(aq)$

Initially 0.260 m 0 0

At equilibrium (0.260 - x) x x

The expression for equilibrium constant for a given reaction is,

$K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

Now put all the given values in this expression, we get

$1.8\times 10^{-5}=\frac{(x)(x)}{(0.260-x)}$

By rearranging the terms, we get the value of 'x'.

$x=2.154\times 10^{-3}m$

Therefore, the equilibrium concentration of $H_3O^+$ at $25^oC$ is, $2.154\times 10^{-3}m$ .

4 0

2 years ago