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vivado [14]
3 years ago
9

In nature, 1.07% of the atoms in a carbon sample are C-13 atoms. Show a numerical setup for calculating the number of C-13 atoms

in a sample containing atoms of carbon.
Chemistry
1 answer:
Vera_Pavlovna [14]3 years ago
6 0
If we let x be the number of sample which contains the atoms of carbon (C). Given that according to study, the amount of atoms in carbon sample that are C-13 atoms is 1.07%, the mathematical set-up that would represent the sample is,
                               = (0.0107)(C)

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Which habitat has salmon with less disease per 100 salmon?
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The molar mass of SO3 is grams. in significant figures
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80.066 g/mol

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What mass of water will change its temperature by 3.0 0C when 525 J of heat is added to it? The specific heat of water is 4.184
pav-90 [236]

Answer:

Mass of water = 41.8 g

Explanation:

Given data:

Mass of water = ?

Change in temperature = 3.0 °C

Specific heat capacity = 4.184 j/g.°C

Heat absorbed = 525 j

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 3.0°C

Now we will put the values in formula.

525 J = m × 4.184 j/g.°C  × 3.0°C

525 J = m × 12.552 j/g

m =  525 J/ 12.552 j/g

m = 41.8 g

4 0
3 years ago
How can you tell whether a fat contains primarily saturated or unsaturated fatty acids?
Romashka-Z-Leto [24]
In terms of physically observing it, if the substance is a lions based substance, like oil or margarine, depending on the state, this will inform how well the hydrocarbons can pack together well or not, or will there be bends in the ring preventing the molecules from packing well.

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3 0
3 years ago
Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

Read more about rate constant

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5 0
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