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ss7ja [257]
3 years ago
14

Mountains and valleys are some of the features found in the layer of rock at Earth's surface called the

Chemistry
1 answer:
loris [4]3 years ago
5 0

Answer:

Narrow

Explanation:

Narrow

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I need A through D, I have no idea on what I'm supposed to do
geniusboy [140]

a. 48.6 is magnesium and 32.0 is oxygen

b. 80.6

c. also 80.6

d. yes, because the product has equal mass to the reactants

3 0
3 years ago
Please please help me with this chemistry question over limiting reactants it’s due by midnight! I’ll mark you BRAINIEST!!!
OverLord2011 [107]

Answer: 1.5 moles

Explanation: one mole Zn uses 2 moles HCl.

1.5 moles Zn uses 3.0 mol HCl. Then Zn is a limiting reactant

And produces equal amount of H2.

4 0
3 years ago
A solid is 5 cm tall, 3 am wide, and 2 cm thick. It has a mass of 135 g. What is the solid's density?
gizmo_the_mogwai [7]

<u>Answer:</u>

density of solid = 4.5 g/cm³

<u>Explanation:</u>

density = mass ÷ volume

• We know the mass of the object (135 g). We need to calculate the volume:

Volume = length × width × thickness

             = 5 cm × 3 cm × 2 cm

             = 30 cm³

• Now we can calculate the solid's density:

density = 135 g ÷ 30 cm³

            = 4.5 g/cm³

4 0
2 years ago
AgCl is..<br> A)an element <br> B)a compound <br> C.)a mixture
SOVA2 [1]
It is an A compound
6 0
2 years ago
Read 2 more answers
Calculate the pH of a solution that is 0.235M benzoic acid and 0.130M sodium benzoate, a salt whose anion is the conjugate base
lesantik [10]

Hello!

We have the following data:

ps: we apply Ka in benzoic acid to the solution.

[acid] = 0.235 M (mol/L)

[salt] = 0.130 M (mol/L)

pKa (acetic acid buffer) =?

pH of a buffer =?

Let us first find pKa of benzoic acid, knowing that Ka (benzoic acid) = 6.20*10^{-5}

So:

pKa = - log\:(Ka)

pKa = - log\:(6.20*10^{-5})

pKa = 5 - log\:6.20

pKa = 5 - 0.79

\boxed{pKa = 4.21}

Now, using the abovementioned data for the pH formula of a buffer solution or (Henderson-Hasselbalch equation), we have:

pH = pKa + log\:\dfrac{[salt]}{[acid]}

pH = 4.21 + log\:\dfrac{0.130}{0.235}

pH = 4.21 + log\:0.55

pH = 4.21 + (-0.26)

pH = 4.21 - 0.26

\boxed{\boxed{pH = 3.95}}\end{array}}\qquad\checkmark

Note:. The pH <7, then we have an acidic solution.

I Hope this helps, greetings ... DexteR! =)

8 0
2 years ago
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