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aleksandrvk [35]
2 years ago
8

How many significant figures are in 8.00 (10)-3

Chemistry
1 answer:
Kay [80]2 years ago
6 0

Answer:  1 significant figure

Explanation:  8.00*10^-3 is simply 1000/8 which will give you 0.008 and in significant numbers do not count zero.

So there you have one significant figure which is the figure 8

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This earth system contains water and warms the atmosphere.
sleet_krkn [62]

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Explanation:

4 0
3 years ago
How much concentrated 18M sulfuric acid is needed to prepare 250mL of a 6.0M solution?
slavikrds [6]
C₁ * V₁ = C₂ * V₂ 

18 * V₁ = 6.0 * 250

18 V₁ = 1500

V₁ = 1500 / 18

V₁ = 83.33 mL

hope this helps!


5 0
3 years ago
3. At STP, how many liters of oxygen gas are required to react completely with 3.2 x 10^22
Mrrafil [7]

The volume of oxygen required to react with 3.2 * 10²² molecules of hydrogen is 0.56 L.

<h3>What is stoichiometric law?</h3>

The stoichiometric law has been given as the representation of the moles of product and reactant in a chemical reaction are represented by the stoichiometric coefficient.

It has been known that 1 mole of a compound has 6.023 * 10²³ molecules. Thus, the moles of hydrogen gas equivalent to 3.2 * 10²² molecules has been:

6.023 * 10²³ molecules = 1 mole

3.2 * 10²² molecules = 1/6.023 * 10²³ * 3.2 * 10²² moles

3.2 * 10²²  molecules = 0.05 moles

Thus, the moles of hydrogen gas available is 0.05 moles.

From the stoichiometric law, according to the balanced chemical equation,

2 moles of hydrogen requires = 1 moles of oxygen

0.05 moles of hydrogen requires = 1/2 * 0.05 moles oxygen

0.05 moles of hydrogen requires =  0.025 moles oxygen

Thus, the moles of oxygen required is 0.025 moles. At STP, a mole of gas been equivalent to 22.4 L . Thus, the volume of 0.025 moles oxygen has been:

1 mole = 22.4 L

0.025 moles = 22.4 * 0.025 L

0.025 mole = 0.56 L

Thus, the volume of oxygen required to react with 3.2 * 10²² molecules of hydrogen is 0.56 L.

Learn more about stoichiometric law, here:

brainly.com/question/23742235

#SPJ1

7 0
2 years ago
What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 5.0? Not
Leni [432]

Answer:

Acetic acid 0,055M and acetate 0,095M.

Explanation:

It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:

pH = pka + log₁₀ [A⁻]/[HA] <em>-Where A⁻ is acetate ion and HA is acetic acid-</em>

Replacing:

5,0 = 4,76 + log₁₀ [A⁻]/[HA]

<em>1,7378 =  [A⁻]/[HA] </em><em>(1)</em>

As concentration of buffer is 0,15M, it is possible to write:

<em>[A⁻] + [HA] = 0,15M </em><em>(2)</em>

Replacing (1) in (2):

1,7378[HA] + [HA] = 0,15M

2,7378[HA] = 0,15M

[HA] = 0,055M

Thus, [A⁻] = 0,095M

That means you need <em>acetic acid 0,055M</em> and <em>acetate 0,095M</em> to obtain the buffer you need.

i hope it helps!

7 0
3 years ago
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