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Ganezh [65]
3 years ago
6

What do you think of my lab report so far?

Chemistry
2 answers:
8090 [49]3 years ago
5 0
Fix ur transition, it sounds choppy

bekas [8.4K]3 years ago
5 0
I can only see one page. If you are doing this under Perdue Owl structure, then you will need a cover page as well as numbers on the page and a reference page at the end. Some professors also want a content page
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How many milliliters of a 5.0 M H2SO4 stock solution would you need to prepare 108.0 mL of 0.45 M H2SO4?
PIT_PIT [208]
For the purpose we will use solution dilution equation:
c1xV1=c2xV2
Where, c1 - concentration of stock solution; V1 - a volume of stock solution needed to make the new solution; c2 - final concentration of new solution; V2 - final volume of new solution.
c1 = 5.00 M
c2 = 0.45 M
V1 = ?
V2 = 108 L
When we plug values into the equation, we get following:
5 x V1 = 0.45 x 108
<span>V1 = </span>9.72 L
7 0
4 years ago
2. What group is the atom in?<br> 2 or 2A<br> 6<br> 14 <br> 1
Juliette [100K]

Answer:

\Large \boxed{\mathrm{Group \ 4}}

Explanation:

The number of valence electrons tells us the group number of the neutral atom.

The atom has 4 valence electrons.

The atom is in group 4.

5 0
3 years ago
Read 2 more answers
30g of naoh is dissolved in 1.5 liter solution the active mass of naoh is?
Pepsi [2]

Answer:

0.5mol/L

Explanation:

First, let us calculate the number of mole NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the question = 30g

Number of mole = Mass /Molar Mass

Number of mole = 30/40 = 0.75mol

Volume = 1.5L

Active mass = mole/Volume

Active mass = 0.75mol/1.5L

Active mass = 0.5mol/L

7 0
3 years ago
In an acid-base neutralization reaction 38.74 ml of 0.500 m potassium hydroxide reacts with 50.00 ml of sulfuric acid solution.
DaniilM [7]

Answer:

0.2 M.

Explanation:

  • For the acid-base neutralization, we have the role:

The no. of millimoles of acid is equal to that of the base at the neutralization.

<em>∴ (XMV) KOH = (XMV) H₂SO₄.</em>

X is the no. of reproducible H⁺ (for acid) or OH⁻ (for base),

M is the molarity.

V is the volume.

  • For KOH:

X = 1, M = 0.5 M, V = 38.74 mL.

  • For H₂SO₄:

X = 2, M = ??? M, V = 50.0 mL.

∴ M of H₂SO₄ = (XMV) KOH/(XV) H₂SO₄ = (1)(0.5 M)(38.74 mL)/(2)(50.0 mL) = 0.1937 M ≅ 0.2 M.

5 0
3 years ago
What causes eclipses? select three options
arlik [135]

Answer:

The first, second, and 4th option

Explanation:

4 0
3 years ago
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