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Stels [109]
1 year ago
10

Aqueous hydrobromic acid reacts with solid sodium hydroxide to produce aqueous sodium bromide and liquid water . What is the the

oretical yield of water formed from the reaction of of hydrobromic acid and of sodium hydroxide
Chemistry
1 answer:
natima [27]1 year ago
8 0

Mass of water produced is equal to 7.2 grams.

The reactants are reacting according to the following equation:

HBr + NaOH = NaBr + H2O

1 mole of HBr requires 1 mole of NaOH to be totally transformed into NaBr and water.

Now, let us calculate the amounts of HBr and NaOH in mols:

m(HBr) = 67.2g

m(NaOH) = 16g

M(HBr) = 80.9g/mol

M(NaOH) = 40g/mol

Following the simple relation

n(moles) = m(grams)/M(g/mol)

n(HBr) = 67.2g/(80.9g/mol)

n(NaOH) = 16g/(40g/mol)

n(H2O) = n(NaOH) = 0.40 mol, which is:

m(H2O) = n(H2O) x M(H2O)

0.40mol∗18g/mol=7.2g

Finally, the mass of water produced is equal to 7.2 grams.

Learn more about Hydrobromic acid here:

brainly.com/question/21544156

#SPJ4

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Answer: 63.26%

Explanation:

If we let the abundance of the first isotope be x, then:

69.66=(68.9255)(x)+(70.9247)(1-x)\\69.66=68.9255x+70.9247-70.9247x\\-1.2647=-1.9992x\\x=\frac{-1.2647}{-1.9992} \approx 0.6326

Which is equal to <u>63.26%</u>

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In this part of the periodic table, what type of elements are in the group that includes elements cu, ag, and au?
tresset_1 [31]
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3 years ago
Antifreeze is mixed with water in a car's cooling system because the solution what?
Nikolay [14]
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8 0
3 years ago
Consider the following reaction 2 N2O(g) =&gt; 2 N2(g) + O2(g) rate = k[N2O]. For an initial concentration of N2O of 0.50 M, cal
den301095 [7]

Answer:

After 2.0 minutes the concentration of N2O is 0.3325 M

Explanation:

Step 1: Data given

rate = k[N2O]

initial concentration of N2O of 0.50 M

k = 3.4 * 10^-3/s

Step 2: The balanced equation

2N2O(g) → 2 N2(g) + O2(g)  

Step 3: Calculate the concentration of N2O after 2.0 minutes

We use the rate law to derive a time dependent equation.

-d[N2O]/dt = k[N2O]

ln[N2O] = -kt + ln[N2O]i

 ⇒ with k = 3.4 *10^-3 /s

⇒ with t = 2.0 minutes = 120s

⇒ with [N2O]i = initial conc of N2O = 0.50 M

ln[N2O] = -(3.4*10^-3/s)*(120s) + ln(0.5)

ln[N2O] = -1.101

e^(ln[N2O]) = e^(-1.1011)

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After 2.0 minutes the concentration of N2O is 0.3325 M

3 0
3 years ago
Help! Please :) Thank you
BartSMP [9]
First one??

I believe this is the correct answer
5 0
2 years ago
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