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Bess [88]
3 years ago
5

Part

Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
7 0

The molar concentration will be greater than 0.01 M KIO_{3}.


Since more of the compound was measured out than what was calculated, you can think of the solution as being 'stronger' than what it was calculated to be. Since a 'stronger' concentration results in a number that is higher, the molarity of this solution is going to be greater than 0.01 M.

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james needs 4.20 moles of nabr for an experiment. how many grams of nabr must he measure out to have 4.20 moles?
kondor19780726 [428]
75.7 grams.
Convert moles to grams using this:

Moles • grams/1 mole = grams

(Grams is over 1 mole)
6 0
3 years ago
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Write the balanced equation for the decomposition of hydrogen peroxide to form water and oxygen.
Lerok [7]
Hydrogen peroxide is H2O2, while water is H2O and oxygen (a diatomic gas) is O2. The (unbalanced) reaction is:
H2O2 --> H2O + O2
Notice that the H2O2 has 2 H atoms, and so does H2. This means that both must have the same coefficients, and we can adjust the coefficient of O2. Since H2O2 has 2 O atoms, and H2O has 1, we multiply O2 by 1/2:
H2O2 --> H2O + (1/2)O2
This has an equivalent number of H and O atoms on either side, but we want the coefficients to be whole numbers, so we multiply everything by 2:
2H2O2 --> 2H2O + O2
7 0
3 years ago
The blue underlined part of the chemical formula above represents a(n) ______________.
Nataly [62]
I’m not totally sure but I think the answer is B
7 0
2 years ago
Question 4 Unsaved
weqwewe [10]
I think it is eutrophication
4 0
3 years ago
Chlorine atoms react with methane, forming HCl and CH3. The rate constant for the reaction was determined to be 3.600×107 at 278
Ratling [72]

<u>Answer:</u> The activation energy for the reaction is 40.143 kJ/mol

<u>Explanation:</u>

To calculate activation energy of the reaction, we use Arrhenius equation for two different temperatures, which is:

\ln(\frac{K_{317K}}{K_{278K}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{317K} = equilibrium constant at 317 K = 3.050\times 10^{8}M^{-1}s^{-1}

K_{278K} = equilibrium constant at 278 K = 3.600\times 10^{7}M^{-1}s^{-1}

E_a = Activation energy = ?

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 278 K

T_2 = final temperature = 317 K

Putting values in above equation, we get:

\ln(\frac{3.050\times 10^8}{3.600\times 10^{7}})=\frac{E_a}{8.314J/mol.K}[\frac{1}{278}-\frac{1}{317}]\\\\E_a=40143.3J/mol=40.143kJ/mol

Hence, the activation energy for the reaction is 40.143 kJ/mol

8 0
3 years ago
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