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Fudgin [204]
3 years ago
15

Convert 10.9 m to mm

Chemistry
1 answer:
wolverine [178]3 years ago
6 0

Explanation:

<h2><em>10900 millimetres</em></h2><h2><em>that's</em><em> all</em></h2><h2><em>thank</em><em> you</em></h2><h2><em>please </em><em>mark</em><em> brainlist</em></h2>
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The three major groups on the peridoic table are the ____
qaws [65]
Metals on the left side, metalloids on the staircase, nonmetals on right side
5 0
3 years ago
If the concentrated form of blood expander is 9.0 mol/L and one of your companions, a nurse, tells you that you need to have a f
I am Lyosha [343]

Answer:

0.032 L or 32 mL

Explanation:

Use the dilution equation M1V1 = M2V2

M1 = 9.0 M

V1 = This is what we're looking for.

M2 = 0.145 M

V2 = 2 L

Solve for V1 --> V1 = M2V2/M1

V1 = (0.145 M)(2 L) / (9.0 M) = 0.032 L

3 0
3 years ago
What does a neutralization reaction produce?
algol13

Answer:

hope this help....

Explanation:

Neutralization is the reaction of an acid and a base, which forms water and a salt. Net ionic equations for neutralization reactions may include solid acids, solid bases, solid salts, and water.

7 0
2 years ago
Read 2 more answers
Calculate difference, ΔH-ΔE=Δ(PV) for the combustion reaction of 1 mole of heptane. (Assume standard state conditions and 298 K
Mamont248 [21]

Answer:

ΔPV = -9911 J

Explanation:

The combustion of 1 mole of heptane is:

C₇H₁₆(l) + 11O₂(g) → 7CO₂(g) + 8H₂O(l)

The change in number of moles of gas molecules is:

Δn = 7 moles products - 11 moles reactants = -<em>4 moles</em>

Using ideal gas ΔPV is:

ΔPV = ΔnRT

Where:

Δn is -4mol

R is 8,314472 J/molK

T is 298K

Replacing:

<em>ΔPV = -9911 J</em>

<em />

I hope it helps!

6 0
3 years ago
A student has a 0.00124 M HCl (aq) solution and she titrates 25.00 mL of this solution against an unknown potassium hydroxide so
malfutka [58]

Answer:

The Molarity of KOH is

7,01.10^{-4}M

Explanation:

The endpoint indicates the volume necessary to neutralize the moles of acid.

In other words, the point at which the moles of both solutions are the same.

M_{(HCl)}V_{HCl}=n\\ \\M_{(KOH)}V_{KOH}=n

we match these equations and find the concentration of KOH

M_{(HCl)}. V_{(HCl)} =M_{(KOH)} .V_{(KOH)}\\ \\M_{(KOH)}=\frac{M_{(HCl)}. V_{(HCl)}}{V_{(KOH)}} \\\\M_{(KOH) =\frac{(25mL)(0,00124m)}{(44,25mL)}\\\\

M_{(KOH)}=7,01.10^{-4}

5 0
3 years ago
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