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GREYUIT [131]
1 year ago
5

an m peak has a relative intensity of 42%. an m 1 peak has a relative intensity of 6.5%. how many carbons are in the molecule?

Chemistry
1 answer:
irakobra [83]1 year ago
5 0

5 carbons are in the molecule. Isotopes are atoms with a constant number of protons but a variable number of neutrons.

<h3>What factors affect an isotope's relative abundance?</h3>

Isotopes are atoms with a constant number of protons but a variable number of neutrons. Atomic masses vary among isotopes.

The proportion of atoms with a particular atomic mass that can be found in a sample of an element that is found naturally is known as the relative abundance of an isotope.

Relative abundance chemistry issues should be solved using the formula below:

(M1)(x) + (M2)(1-x) = M (E)

Issue illustration Determine the relative abundance of the isotopes if the masses of two different nitrogen isotopes are 0.42 amu for carbon-0.42 and 0.065 amu forcarbon-0.065.

Equation x = 0.996 can be found using algebra.

0.42 * 0.996 + 0.065 *0.996 = 0.424794.

5 carbons are in the molecule.  

To learn more about Relative abundance of an isotope refer to:

brainly.com/question/24873591

#SPJ4

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A 1.555-g sample of baking soda decomposes with heat to produce 0.991 g Na2CO3. Refer to Example Exercise 14.l and show the calc
Sunny_sXe [5.5K]

Answer:

a) 101%

b)59.7%

Explanation:

The equation for the thermal decomposition of baking soda is shown;

2NaHCO3 → Na2CO3 + H2O + CO2

Number of moles of baking soda= mass/molar mass= 1.555g/84.007 g/mol = 0.0185 moles

From the reaction equation;

2 moles of baking soda yields 1 mole of sodium carbonate

0.0185 moles of baking soda will yield = 0.0185 moles ×1 /2 = 9.25 ×10^-3 moles of sodium carbonate.

Therefore, mass of sodium carbonate= 9.25 ×10^-3 moles × 106gmol-1= 0.9805 g of sodium carbonate. This is the theoretical yield of sodium carbonate.

%yield = actual yield/theoretical yield ×100

% yield = 0.991/0.9805 ×100

%yield = 101%

Since ;

2NaHCO3 → Na2CO3 + H2O + CO2

And H2O + CO2 ---> H2CO3

Hence I can write, 2NaHCO3 → Na2CO3 + H2CO3

Molar mass of H2CO3= 62.03 gmol-1

Molar mass of baking soda= 84 gmol-1

Therefore, mass of baking soda=

0.325/62.03 × 2 × 84 = 0.88 g of NaHCO3

% of NaHCO3= 0.88/1.473 × 100 = 59.7%

7 0
3 years ago
The voltage generated by the zinc concentration cell described by the line notation Zn ( s ) ∣ ∣ Zn 2 + ( aq , 0.100 M ) ∥ ∥ Zn
Oxana [17]

Answer:

0.193 M

Explanation:

We need to calculate the Zn²⁺ concentration at the cathode where reduction occurs which is the right side in the expression:

Zn2+(aq,0.100 M) ‖ Zn2+(aq,? M) | Zn(s)

Zn²⁺ (aq,?) + 2 e⁻   ⇒ Zn (s)

and oxidation will occur in the anode

Zn (s) ⇒ Zn²⁺ (aq, 0.100 M ) + 2 e⁻

and the overall reaction is

Zn²⁺ (aq,?) ⇒ Zn²⁺ (0.100 M )

The driving force is the difference in concentration and E the electromotive force will be given by

E = Eº - 0.0592/2 log [Zn²⁺ (0.100 M) / Zn²⁺ (M) ]

Plugging the value for E and knowing Eº is cero because we have the same electrodes, we have

17.0 x 10⁻³ = 0 - 0.0592 log 0.100/ X =

- 17.0 x 10⁻³ / 0.0592 = log 0.100 / X

- 0.287 = log (0.100 / X)

Taking inverse log to both sides of the equation

0.516 = 0.100 / X   ⇒ X = 0.100 / 0.516 = 0.193 M

4 0
3 years ago
PLEASE HELP FAST !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
MakcuM [25]
The answer is D: 400N
your welcome

5 0
3 years ago
HELP PLS 50 POINTSSSS
Alenkasestr [34]

Answer:

Carbon dioxide is a product (cellular respiration)

Carbon dioxide is a reactant(photosynthesis)

Carried out in animals(cellular respiration)

Carried out in plants(both)

Chemical reaction(cellular respiration)

Oxygen is a product(photosynthesis)

Oxygen is a reactant(cellular respiration)

Produces usable energy source(photosynthesis)

Your welcome!!! Plz mark brainliest.

4 0
3 years ago
Read 2 more answers
2. HOW MUCH HEAT IS REQUIRED TO BE RELEASED WHEN
jarptica [38.1K]

Answer: -112200J

Explanation:

The amount of heat (Q) released from an heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = ?

Mass of water vapour = 30.0g

C = 187 J/ G°C

Φ = (Final temperature - Initial temperature)

= 100°C - 120°C = -20°C

Then apply the formula, Q = MCΦ

Q = 30.0g x 187 J/ G°C x -20°C

Q = -112200J (The negative sign does indicates that heat was released to the surroundings)

Thus, -112200 joules of heat is released when cooling the superheated vapour.

5 0
3 years ago
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