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2 years ago

If i have 340mL of a 1.5 M NaBr solution, What will the concentration be for 1000mL?

Chemistry

1 answer:

dlinn [17]2 years ago

3 0

Answer:

0.51M

Explanation:

Given parameters:

Initial volume of NaBr = 340mL

Initial molarity = 1.5M

Final volume = 1000mL

Unknown:

Final molarity = ?

Solution;

This is a dilution problem whereas the concentration of a compound changes from one to another.

In this kind of problem, we must establish that the number of moles still remains the same.

number of moles initially before diluting = number of moles after dilution

Number of moles = Molarity x volume

Let us find the number of moles;

Number of moles = initial volume x initial molarity

Convert mL to dm³;

1000mL = 1dm³

340mL gives $\frac{340}{1000}$ = 0.34dm³

Number of moles = initial volume x initial molarity = 0.34 x 1.5 = 0.51moles

Now to find the new molarity/concentration;

Final molarity = $\frac{number of moles}{Volume}$ = $\frac{0.51}{1}$ = 0.51M

We can see a massive drop in molarity this is due to dilution of the initial concentration.

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A student is performing a titration to determine the concentration of a 20.0 mL sample of hydrochloric acid. What did the studen

irakobra [83]

Answer:

0.25M HCl

Explanation:

The reaction of HCl with NaOH is:

HCl + NaOH ⇄ H₂O + NaCl

Where 1 mole of HCl reacts per mole of NaOH

The end point was reached when the student added:

0.0500L × (0.1mol / L) = 0.00500 moles of NaOH

As 1 mole of HCl reacted per mole of NaOH, moles of HCl present are:

0.00500 moles HCl

The volume of the sample of hydrochloric acid was 20.0mL = 0.0200L, and concentration of the sample is:

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2 years ago

How do the energy and the most probable location of an electron in the third shell of an atom compare to the energy and the most

solmaris [256]

The answer is (2). You can think about this question in terms of the Bohr's model of the atom or in terms of quantum chemistry. In the Bohr model, electrons exist in discrete "shells," each respresenting a fixed spherical distance from the nucleus in which electrons of certain energy levels orbit the nucleus. The larger the shell (the greater the "orbit" radius), the greater the energy of the "orbiting" electron (I use quotations because electrons don't actually orbit the nucleus in the traditional sense, as you may know). Thus, according to the Bohr model, a third shell electron should be farther from the nucleus and have greater energy than an electron in the first shell. The quantum model is differs drastically from the Bohr model in many ways, but the essence is the same. A larger principal quantum number indicates 1) greater overall energy and 2) a probability distribution spread a bit more outward.

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2 years ago

Determine the mass of carbon iv oxide ,produced on burning 104g of ethyne

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163 grams (3 sig. fig.).

<h3>Explanation</h3>

Formula of carbon(IV) oxide (a.k.a. carbon dioxide): $\text{CO}_2$ .
Molar mass of $\text{CO}_2$ : $\underbrace{12.01}_{\text{C}} + 2\times\underbrace{16.00}_{\text{O}}=44.01\;\text{g}\cdot\text{mol}^{-1}$ .

Formula of ethyne: structural $\text{H}-\text{C}\equiv\text{C}-\text{H}$ or molecular $\text{C}_2\text{H}_2$ .
Molar mass of $\text{C}_2\text{H}_2$ : $2\times\underbrace{12.01}_{\text{C}}+2 \times\underbrace{16.00}_{\text{O}} = 56.02\;\text{g}\cdot\text{mol}^{-1}$ .

All carbon atoms in that 104 grams of ethyne will end up in $\text{CO}_2$ . Number of moles of molecules in 104 grams of ethyne:

$n = \dfrac{m}{M} = \dfrac{104}{56.02} = 1.85648\;\text{mol}$ .

There are two carbon atoms in each ethyne molecule. Number of carbon atoms in that many ethyne molecules:

$n(\text{C}) = 2\;n(\text{C}_2\text{H}_2) = 3.71296\;\text{mol}$ .

There are one carbon atom in each $\text{CO}_2$ molecule. In case oxygen is in excess, all those carbon atoms from that 104 grams of ethyne will make $n(\text{CO}_2) = n(\text{C}) =3.71296\;\text{mol}$ of $\text{CO}_2$ .

Mass of all those $\text{CO}_2$ molecules:

$m = n\cdot M = 163\;\text{g}$ . (3 sig. fig. as in the mass of ethyne.)

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3 years ago

You have 4.5 x 10^24 particles of C3H8. How many grams are present?

Citrus2011 [14]

Answer:

33 g.

Explanation:

Hello there!

In this case, for these particle-mole-mass relationships problems, it is necessary for us to recall the following equivalence statement, based off the molar mass of the involved compound, C3H8, one mole of particles and the Avogadro's number:

$1mol=44.11g=6.022x10^{23}molecules$

In such a way, we can set up the following expression for the calculation of the mass in the given particles of propane:

$4.5x10^{23}molecules*\frac{1mol}{6.022x10^{23}molecules} *\frac{44.11g}{1mol}\\\\33g$

Best regards!

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2 years ago

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Answer:

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2 years ago