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3 years ago

8

The isotope of carbon used in archaeological dating is 14^6C . How many protons, neutrons, and electrons does an atom of 14^6C h

ave?

1 answer:

vampirchik [111]3 years ago

8 0

Answer:

6

8

6

Explanation:

Isotope given:

¹⁴₆C

In specie written as this;

Superscript = Mass number

Subscript = Atomic number

To find the protons, it is the same as the atomic number;

Protons = Atomic number = 6

Neutrons have no charges;

Neutrons = Mass number - Atomic number =

Neutrons = 14 - 6 = 8

The number of electrons is the same as the atomic number = 6

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What is the term for the amount of energy that is needed for a chemical reaction to occur?

cupoosta [38]

<span>This would be the activation energy. This is usually in the form of heat, which allows the reaction to undergo some sort of transition. Many times, enzymes can be used as catalysts to lower the activation energy required for the reaction to take place.</span>

3 0

3 years ago

A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so

Xelga [282]

Answer : The equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

Explanation :

First we have to calculate the initial moles of $Fe^{3+}$ and $SCN^-$ .

$\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}$

$\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol$

and,

$\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}$

$\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol$

The given balanced chemical reaction is,

$Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$

Since 1 mole of $Fe^{3+}$ reacts with 1 mole of $SCN^-$ to give 1 mole of $FeSCN^{2+}$

The limiting reagent is, $SCN^-$

So, the number of moles of $FeSCN^{2+}$ = 0.0020 mmole

Now we have to calculate the concentration of $FeSCN^{2+}$ .

$\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M$

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution

l = path length

$\epsilon$ = molar absorptivity coefficient

$\epsilon$ and l are same for stock solution and dilute solution. So,

$\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}$

For trial solution:

The equilibrium concentration of $SCN^-$ is,

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{initial}$ = 0.00050 M

Now calculate the $[FeSCN^{2+}]$ .

$C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M$

Now calculate the concentration of $SCN^-$ .

$[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]$

$[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)$

$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

Therefore, the equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

5 0

3 years ago

According to VSEPR theory, a molecule with the general formula AX2E2 (A is the central atom, X is the bonded atom and E is a lon

ratelena [41]

Answer:

see explanation

Explanation:

An AX₂E₂ geometry is derived from an AX₄ parent geometry and is based upon 4 regions of electron density about the central element and defines a tetrahedral geometry and the geometry is bent angular.

An example is the water molecule (H₂O) with two covalent O - H bonds and two free pair electrons on the central oxygen element.

4 0

3 years ago

What about 50 g of water?<br> I need help what this

ELEN [110]

Answer:

3.38 Tablespoons

10.14 Teaspoons

0.21 U.S. Cups

0.18 Imperial Cups

0.20 Metric Cups

50.00 Milliliters

Explanation:

3 0

3 years ago

The rate constant for a certain reaction is measured at two different temperatures:

Talja [164]

Answer: The activation energy Ea for this reaction is 22689.8 J/mol

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

$ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

$k_1$ = rate constant at temperature $T_1$ = $2.3\times 10^8$

$k_2$ = rate constant at temperature $T_2$ = $4.8\times 10^8$

$E_a$ = activation energy = ?

R= gas constant = 8.314 J/kmol

$T_1$ = temperature = $280.0^0C=(273+280)=553K$

$T_2$ = temperature = $376.0^0C=(273+376)=649K$

Putting in the values ::

$ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]$

$E_a=22689.8J/mol$

The activation energy Ea for this reaction is 22689.8 J/mol

3 0

3 years ago