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Iteru [2.4K]
4 years ago
7

2 ClO2 (aq) + 2OH- (aq)→ ClO3- (aq) + ClO2- + H2O (l) was studied with the following results: Experiment [ClO2] (M) [OH-] (M) In

itial Rate (M/s) 1 0.060 0.030 0.0248 2 0.020 0.030 0.00276 3 0.020 0.090 0.00828 a. Determine the rate law for the reaction. b. Calculate the value of the rate constant with the proper units. c. Calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M.
Chemistry
1 answer:
erica [24]4 years ago
3 0

Explanation:

2 ClO2 (aq) + 2OH- (aq)→ ClO3- (aq) + ClO2- + H2O (l)

The data is given as;

Experiment [ClO2] (M) [OH-] (M) Initial Rate (M/s)

1 0.060 0.030 0.0248

2 0.020 0.030 0.00276

3 0.020 0.090 0.00828

a) Rate law is given as;

Rate = k [ClO2]^x [OH-]^y

From  experiments 2 and 3, tripling the concentration of  [OH-] also triples the rate of the reaction. This means the reaction  is first order with respect to  [OH-]

From experiments 1 and 2, when the [ClO2] decreases by a factor of 3, the rate decreases by a factor of 9. This means the reaction is second order with respect to [ClO2]

Rate =  k [ClO2]² [OH-]

b. Calculate the value of the rate constant with the proper units.

Taking experiment 1,

0.0248 = k (0.060)²(0.030)

k = 0.0248 / 0.000108

k = 229.63 M-2 s-1

c. Calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M.

Rate =  229.63  [ClO2]² [OH-]

Rate = 229.63 (0.100)²(0.050)

Rate = 0.1148 M/s

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Calculate the number of grams of deuterium in an 80,000-L swimming pool, given deuterium is 0.0150% of natural hydrogen
fiasKO [112]

Answer:

m_{Deu}=2666.7g

Explanation:

Hello,

In this case, we can apply the following mole-mass-density relationship in order to compute the required grams of deuterium, considering that it is the 0.0150% (molar basis) of natural hydrogen (H₂):

m_{Deu}=80000LH_2O*\frac{1000gH_2O}{1LH_2O}*\frac{1molH_2O}{18gH_2O}   *\frac{1molH_2}{1molH_2O} *\frac{2gH_2}{1molH_2} *\frac{2*0.0150g\ Deu}{100gH_2} \\\\m_{Deu}=2666.7g

Best regards.

4 0
3 years ago
In a laboratory setting, chemists strive to be
Kitty [74]
Accurate and precise
3 0
4 years ago
1. When you have 43 grams of sodium, how many grams of NaCl can you produce?
seropon [69]

Molar mass is the amount grams that one mole weighs.

Explanation: You need to find the molar mass of NaCl which is the same as the amu on the periodic table in grams. So it is 22.99(Na) + 35.45(Cl) = 58.44

You also know that for every mole of NaCl you have 1 mole of Na because every molecule of NaCl has 1 atom of Na.

Finally, using the periodic table, again, you see that the molar mass of Na is 22.99.

Then using stoichiometry, you can find the grams of sodium.

100(g NaCl) * 1 mol (NaCl)/58.44 g (NaCl) * 1 mol (Na)/ 1 mol (NaCl) * 22.99 (g of Na)/ 1 mol (Na)

which equals 39.339435 g of Na.

If you need to maintain significant figures the answer will be 40.

hope this helps

6 0
3 years ago
Determine the number of charged particles in nucleus of calcium atom then deduce the number of electrons
Dafna1 [17]

Answer:

detail is given below.

Explanation:

The charged particles of nucleus are protons while neutrons are neutral having no charge.

We know that an atom consist of electrons, protons and neutrons. Neutrons and protons are present inside the nucleus while electrons are present out side the nucleus.

Electron has a negative charge and is written as e⁻. The mass of electron is 9.10938356×10⁻³¹ Kg . While mass of proton and neutron is 1.672623×10⁻²⁷Kg and 1.674929×10⁻²⁷ Kg respectively.

Symbol of proton= P⁺  

Symbol of neutron= n⁰  

The number of electron or number of protons are called atomic number while mass number of an atom is sum of protons and neutrons.

one proton contribute 1 amu to the total weight. There are 20 protons and 20 neutrons in Ca thus its atomic mass is 40 amu.

While the atomic number is 20.

3 0
3 years ago
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o-na [289]

Answer:

compound

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MgO is an inorganic compund :) The simplest explanation is that it's not an element b/c there's two elements (Mg & O) in the substance, and it's not a mixture b/c you can't easily separate the two elements.

6 0
3 years ago
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