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scoundrel [369]
4 years ago
5

If the smallest markings on a ruler are one tenth (1/10) of a centimeter, which of the following measurements could be recorded

using that ruler?
A. 2.5 cm
B. 2.52 cm
C. 2.520 cm
D. 2.5200 cm
Chemistry
2 answers:
makkiz [27]4 years ago
4 0
I think it should be A 
Alenkinab [10]4 years ago
4 0

Answer:

The correct answer is the option: A. 2.5 cm

Explanation:

Hello! Let's solve this!

In a rule we see that it is divided into centimeters, they can be 15 centimeters, 20 centimeters, etc.

It is very easy to see the two centimeters, which are the largest lines. To see after two, we count the smallest stripes and we can reach 2.5 centimeters. But no more than that.

We conclude that the correct answer is the option: A. 2.5 cm

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We can drink water straight from the ocean without purifying it. <br><br><br><br> True<br><br> False
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50 POINTS..!!! NEED RHE ANSWER ASAP..!!! consider the original wave at the top of the illustration. compare the original wave to
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4 0
3 years ago
Balance each reaction and write its reaction quotient, Qc:(b) POCl₃(g) ⇄ PCl₃(g) + O₂(g)
Komok [63]

When we balance the given equation

      POCl₃(g) ⇄ PCl₃(g) + O₂(g)

We will get

      2POCl₃(g) ⇄ 2PCl₃(g) + O₂(g)

Solution:

Balancing the given equation

      POCl₃(g) ⇄ PCl₃(g) + O₂(g)

Balancing the number of O

      2POCl₃(g) ⇄ PCl₃(g) + O₂(g)

Balancing the number of P and Cl

      2POCl₃(g) ⇄ 2PCl₃(g) + O₂(g)

We get the balanced equation

      2POCl₃(g) ⇄ 2PCl₃(g) + O₂(g)

The reaction quotient will be

     Qc = [product] / [reactant]

     Qc ​= [PCl₃(g) + O₂(g)] / [POCl₃(g) ]

To learn more click the given link

brainly.com/question/26227625

#SPJ4

4 0
1 year ago
I NEED HELP PLEASE, THANKS! :)
krok68 [10]

Answer:

\large \boxed{\text{2.20 g Pb}}

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       239.27   32.00        207.2

            2PbS   +   3O₂   ⟶  2Pb   +   2SO₃

m/g:      2.54        1.88

2. Calculate the moles of each reactant

\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

4. Calculate the mass of Pb

\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}

5 0
3 years ago
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