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Reil [10]
3 years ago
13

(04.04 LC)

Chemistry
1 answer:
Charra [1.4K]3 years ago
8 0

Answer:

false

Explanation:

it happens in light-independent reactions

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How many molecules are in the quantities below.? 2.0 moles
matrenka [14]
Mole=number of molecules/6.02x10²³
mole=2
number of molecules= 2x6.02x10²³
number of molecules=12.04x10²³
5 0
3 years ago
The incredible catalytic power of enzymes can perhaps best be appreciated by imagining how challenging life would be without jus
ioda

Answer:

t = 7.58 * 10¹⁹ seconds

Explanation:

First order rate constant is given as,

k =  (2.303 /t) log  [A₀] /[Aₙ]

where  [A₀]  is the initial concentraion of the reactant; [Aₙ] is the concentration of the reactant at time, <em>t</em>

[A₀]  = 615 calories;

[Aₙ] = 615 - 480 = 135 calories

k = 2.00 * 10⁻²⁰ sec⁻¹

substituting the values in the equation of the rate constant;

2.00 * 10⁻²⁰ sec⁻¹ = (2.303/t) log (615/135)

(2.00 * 10⁻²⁰ sec⁻¹) / log (615/135) = (2.303/t)

t = 2.303 / 3.037 * 10⁻²⁰

t = 7.58 * 10¹⁹ seconds

8 0
3 years ago
Which Chemical Reaction is balanced?
cupoosta [38]

Answer:

A

Explanation:

The letter A is the correct answer

7 0
3 years ago
An element has two naturally-occurring isotopes. The mass numbers of these isotopes are 113 amu and 115 amu, with natural abunda
DENIUS [597]

Answer: Its average atomic mass is 114.9 amu

Explanation:

Mass of isotope 1 = 113 amu

% abundance of isotope 1 = 5% = \frac{5}{100}=0.05

Mass of isotope 2 = 115 amu

% abundance of isotope 2 = 95% = \frac{95}{100}=0.95

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(113\times 0.05)+(115\times 0.95)]

A=114.9amu

Thus its average atomic mass is 114.9 amu

5 0
3 years ago
Roundup, an herbicide manufactured by Monsanto, has the formula C₃H₈NO₅P. How many moles of molecules are there in a 304.3-g sam
jekas [21]
<span>The molar mass is 169.09

304.3g/169.09g = 1.799mol which rounds to 1.800 mol</span>
7 0
3 years ago
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