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liq [111]
2 years ago
12

A certain element consists of two stable isotopes. The first has atomic mass of 7.02 amu and a percent natural abundance of 92.6

%. The second has an atomic mass of 6.02 amu and a percent natural abundance of 7.42%. What is the atomic weight of the element? in amu
Chemistry
1 answer:
statuscvo [17]2 years ago
8 0

<u>Answer:</u> The average atomic mass of the element is 6.95 amu

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i .....(1)

We are given:

Mass of isotope 1 = 7.02 amu

Percentage abundance of isotope 1 = 92.6 %

Fractional abundance of isotope 1 = 0.926

Mass of isotope 2 = 6.02 amu

Percentage abundance of isotope 2 = 7.42 %

Fractional abundance of isotope 2 = 0.0742

Putting values in equation 1, we get:

\text{Average atomic mass of element}=[(7.02\times 0.926)+(6.02\times 0.0742)]

\text{Average atomic mass of element}=6.95amu

Hence, the average atomic mass of the element is 6.95 amu

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Give a possible explanation for the relative amounts of the isometric methyl nitrobenzoates formed in the nitration reaction. Co
vovikov84 [41]

Answer:

The electrophilic aromatic substitution reaction nitration is used to nitrate methyl benzoate and acetanilide with a nitronium ion. Crystallization was used to purify the product. The melting point was used to determine its purity and the regiochemistry of the products.

Explanation:

Methyl m-Nitrobenzoate is formed in this

reaction rather that ortho/para isomers

because of the ester group of your starting

product of methylbenzoate. The functional

group of ester is a electron withdrawing group

causing nitrobenzene (N02) to become in the

meta position. Thus N02 is a deactivating

group causing itself to be a meta director.

Basically you must look at the substituents

that are attached to your starting benzene ring

in order to figure out whether your reaction

with be ortho/para directors or meta

directors. If the substituents are electron

withdrawing groups then you will be left with

meta as your product but if your substituents

are electron donating groups then your

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4 0
2 years ago
7. Two teams are competing in a tug-of-war contest. Team A is pulling at 4000N and Team B is pulling at 4900N is the opposite di
Leto [7]
Basically team B would win since it is exerting a force of 900N unlike team A ( you can tell by doing 4900N minus 4000N ). It is very unbalanced.

5 0
2 years ago
Read 2 more answers
A student pours exactly 26.9 mL of HCl acid of unknown molarity into a beaker. The student then adds 2 drops of the indicator an
Assoli18 [71]
a.
Acids react with bases and give salt and water and the products.

Hence, HCl reacts with NaOH and gives NaCl salt and H₂O as the products. The reaction is,
            HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

To balance the reaction equation, both sides hould have same number of elements.

Left hand side,                                             Right hand side,
             
H atoms = 2                                               H atoms = 2
            Cl atoms = 1                                               Cl atoms = 1
            Na atoms = 1                                               Na atoms = 1 
           O atoms = 1                                                   O atoms = 1

Hence, the reaction equation is already balanced.

b. 
Molarity (M)= moles of solute (mol) / Volume of the solution (L)
 
          HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Molarity of NaOH = <span>0.13 M
</span>Volume of NaOH added = <span>43.7 mL
Hence, moles of NaOH added = 0.13 M x 43.7 x 10</span>⁻³ L
                                                 = 5.681 x 10⁻³ mol

Stoichiometric ratio between NaOH and HCl is 1 : 1

Hence, moles of HCl = moles of NaOH
                                    = 
5.681 x 10⁻³ mol

5.681 x 10⁻³ mol of HCl was in <span>26.9 mL.

Hence, molarity of HCl = </span>5.681 x 10⁻³ mol / 26.9 x 10⁻³ L
                                     = 0.21 M
6 0
2 years ago
The solubility of N2 in blood can be a serious problem (the "bends") for divers breathing compressed air (78% N2 by volume) at d
OLga [1]

Answer:

The volume is 19.7 mL

Explanation:

<u>Step 1</u>: Given data

Pressure at sea level = 1.00 atm

Pressure at 50 ft = 2.47535 atm

kH for N2 in water at 25°C is 7.0 × 10−4 mol/L·atm

Molarity (M) = kH x P

<u>Step 2</u>: Calculate molarity

M at sea level:

M = 7.0*10^-4 * (1.00atm * 0.78) = 5.46*10^-4 mol/L

M at 50ft:

M = 7.0*10^-4 * (2.47535atm * 0.78) = 13.5*10^-4 mol/L

We should find the volume of N2. To find the volume whe have to find the number of moles first. This we calculate by calculating the difference between M at 50 ft and M at sea level.

13.5*10^-4 mol/L - 5.46*10^-4 mol/L = 8.04*10^-4 mol/L

Step 3: Calculate volume

P*V=nRT

with P = 1.00 atm

with V = TO BE DETERMINED

with n =  8.04*10^-4 mol/L  *1L = 8.04*10^-4

with R= 0.0821 atm * L/ mol *K

with T = 25 °C = 273+25 = 298 Kelvin

To find the volume, we re-organize the formula to: V=nRT/P

V= (8.04*10^-4 mol * 0.0821 (atm*L)/(mol*K)* 298K ) / 1.00atm = 0.0197L = 19.7ml

The volume is 19.7 mL

5 0
2 years ago
Gasoline is a mixture of hydrocarbons, a major component of which is octane, CH3CH2CH2CH2CH2CH2CH2CH3. Octane has a vapor pressu
Nitella [24]

Answer:

\Delta \:H_{vap}=40383.88\ J/mol

Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c

Where,  

P is the vapor pressure

ΔHvap  is the Enthalpy of Vaporization

R is the gas constant (8.314×10⁻³ kJ /mol K)

c is the constant.

For two situations and phases, the equation becomes:

\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)

Given:

P_1 = 13.95 torr

P_2 = 144.78 torr

T_1 = 25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K

T_1 = 298.15 K

T_2 = 75°C  = 348.15 K

So,

\ln \:\left(\:\frac{13.95}{144.78}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314}\:\left(\:\frac{1}{348.15}-\:\frac{1}{298.15}\:\right)

\Delta \:H_{vap}=\ln \left(\frac{13.95}{144.78}\right)\frac{8.314}{\left(\frac{1}{348.15}-\frac{1}{298.15}\right)}

\Delta \:H_{vap}=\frac{8.314}{\frac{1}{348.15}-\frac{1}{298.15}}\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)

\Delta \:H_{vap}=\left(-\frac{863000.86966\dots }{50}\right)\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)

\Delta \:H_{vap}=40383.88\ J/mol

4 0
2 years ago
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