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docker41 [41]
3 years ago
13

What is the volume of .750 M hydrochloride acid required to react completely with 25.00 mL of .290 M NaOH solution?

Chemistry
2 answers:
Whitepunk [10]3 years ago
8 0

Explanation:

lease please please please please please please

crimeas [40]3 years ago
7 0

from

HCl + NaOH----> NaCl +H2O

n of acid=1

n of base=1

Molarity of acid=750M

Molarity of base=290M

volume of base=25ml

from

V<u>a</u><u>=</u><u>M</u><u>b</u><u> </u><u>×</u><u>v</u><u>b</u><u>×</u><u>n</u><u>a</u>

Ma×nb

Va=<u>2</u><u>9</u><u>0</u><u>×</u><u>2</u><u>5</u><u>×</u><u>1</u>

750×1

Va=<u>7</u><u>2</u><u>5</u><u>0</u>

750

Va=9.67ml

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LiOH+HBr---&gt; LiBr +h20 If you start with 10.0 grams of lithium hydroxide, how many grams of lithium bromide will be produced?
kicyunya [14]
<span>LiOH+HBr---> LiBr +h20. Moles of LiOH = 10/24 = 0.41moles. According to stoichiometry, moles of LiOH = moles of LiBr = 0.41moles. Therefore mass of LiBr =moles of LiBr x molecular weight of LiBr = o.41 x 87 = 35.67g. Hope it helps </span>
7 0
3 years ago
Find the number of grams in 16.95 mol hydrogen peroxide (H2O2). Round your
Feliz [49]

Answer: There are 576.46 number of grams present in 16.95 mol hydrogen peroxide (H_{2}O_{2}).

Explanation:

Number of moles is defined as the mass of substance divided by its molar mass.

The molar mass of H_{2}O_{2} is 34.01 g/mol. Hence, mass of hydrogen peroxide present in 16.95 moles is calculated as follows.

Moles = \frac{mass}{molarmass}\\16.95 mol = \frac{mass}{34.01 g/mol}\\mass = 576.46 g

Thus, we can conclude that there are 576.46 number of grams present in 16.95 mol hydrogen peroxide (H_{2}O_{2}).

3 0
3 years ago
Chlorine has two stable isotopes, Cl-35 and Cl-37. If their exact masses are 34.9689 amu and 36.9695 amu, respectively, what is
natima [27]

Answer:

X(Cl-35) = 75.95% => Answer 'A'

Explanation:

34.9689·X(Cl-35) + 36.9695·X(Cl-37) = 35.45; X = fractional abundance

X(Cl-35) + X(Cl-37) = 1  ⇒  X(Cl-37) = 1 - X(Cl-25)

34.9689·X(Cl-35) + 36.9695(1 - X(Cl-35)) = 35.45

34.9689·X(Cl-35) + 36.9695 - 36.9695·X(Cl-35) = 35.45

Rearrange ...

36.9695·X(Cl-35) - 34.9689·X(Cl-35) = 36.9689 - 35.45

2.0006·X(Cl-35) = 1.5195

X(Cl-35) = 1.5195/2.0006 = 0.7595 fractional abundance

⇒ % abundance = 75.95%

3 0
3 years ago
Read 2 more answers
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