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4 years ago

The metric system is based on what number?

Chemistry

1 answer:

Dmitriy789 [7]4 years ago

3 0

Answer:

Explanation:

The prefixes attached to metric units carry the same meaning for all base units. The metric system is based upon powers of ten, which is convenient because: A measurement in the metric system that is represented by a rational number remains a rational number after metric unit conversion.

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In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewher

trasher [3.6K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of $BC_2$ that should used originally is $C_Z_o = 0.4492M$

Explanation:

From the question we are told that

The necessary elementary step is

$2BC_2 ----->4C + B_2$

The time taken for sixth of 0.5 M of reactant to react $t = 9 hr$

The time available is $t_a = 3.5 hr$

The desired concentration to remain $C = 0.42M$

Let Z be the reactant , Y be the first product and X the second product

Generally the elementary rate law is mathematically as

$-r_Z = kC_Z^2 = - \frac{d C_Z}{dt}$

Where k is the rate constant , $C_Z$ is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

For second order reaction

$\frac{1}{C_Z} - \frac{1}{C_Z_o} = kt$

Where $C_Z_o$ is the initial concentration of Z which a value of $C_Z_o = 0.5M$

From the question we are told that it take 9 hours for the concentration of the reactant to become

$C_Z = C_Z_o - \frac{1}{6} C_Z_o$

$C_Z = 0.5 - \frac{0.5}{6}$

$= 0.4167 M$

$\frac{1}{0.4167} - \frac{1}{0.50} = 9 k$

$0.400 = 9 k$

=> $k = 0.044\ L/ mol \cdot hr^{-1}$

For $C_Z = 0.42M$

$\frac{1}{0.42} - \frac{1}{C_Z_o} = 3.5 * 0.044$

$2.38 - 0.154 = \frac{1}{C_Z_o}$

$2.226 = \frac{1}{C_Z_o}$

$C_Z_o = \frac{1}{2.226}$

$C_Z_o = 0.4492M$

4 0

3 years ago

PLEASE HELP----

balandron [24]

Actual yield of Fe2(So4)3 = 18.5g

2FePo4 + 3Na2SO4 -> Fe2(SO4)3 + 2Na3PO4

Mole of FePO4 = mass of it / its molar mass =
25 g / (55.8 + 31 + 16*4) = 0.166 mol

every 2 mole of FePO4 will form 1 mole of Fe2(SO4)3

Mole of Fe2(SO4)3 produced = 0.166 / 2 = 0.0829 mol

0.0829 * (55.8*2 + 3*(32.1+ 16*4)) = 33.148 g of Fe2(SO4)3

18.5 / 33.148 * 100 = 55.8%

3 0

3 years ago

Which of the following elements is a nonmetal? a. selenium (Se) b. yttrium (Y) c. barium (Ba) d. calcium (Ca) e. All of these ar

san4es73 [151]

<u>Answer:</u>

The correct answer option is a. Selenium (Se).

<u>Explanation:</u>

From the given answer options, Selenium (Se) is the only element which is a non metal.

It has the atomic number 34 with atomic weight 78.96. It is a member of the sulfur group of the non metallic elements and falls in period 4 of the Periodic table.

Selenium has non metallic properties which are intermediate between the elements that lie above and below it in the Periodic table.

3 0

3 years ago

Read 2 more answers

A student made a graph to show the chemical equilibrium position of a reaction. The student forgot to label the y-axis of the gr

Kazeer [188]

Answer:

D. concentration, as the concentrations of reactants and products remain unchanged after equilibrium is reached.

Explanation:

I put it for the test and i got right hehe

7 0

3 years ago

What is the atomic weight of a hypothetical element consisting of two isotopes, one with a mass of 62.2 u which is 24% abundant,

Alona [7]

Answer:

The atomic weight of hypothetical element will be 63.568 amu.

Explanation:

Given data:

First isotope mass = 62.2 amu

Percentage abundance of first isotope = 24%

Mass of second isotope = 64 amu

Percentage abundance of second isotope = 100- 24 = 76%

Solution:

The atomic weight of hypothetical element will be the average atomic mass of its isotopes.

Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100

Now we will put the values in formula.

Average atomic mass = [24 ×62.2] + [76× 64] / 100

Average atomic mass = 1492.8 + 4864 / 100

Average atomic mass = 6356.8/100

Average atomic mass = 63.568 amu

3 0

4 years ago