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deff fn [24]
3 years ago
8

1 How many moles of solute are in:

Chemistry
1 answer:
Sonbull [250]3 years ago
5 0

Moles of solute for both a and b are the same = 1 mol

<h3>Further explanation</h3>

Given

a 500 cm³ of solution, of concentration 2 mol/dm³

b 2 litres of solution, of concentration 0.5 mol/dm³

Required

moles of solute

Solution

Molarity shows the number of moles of solute in every 1 liter of solution or mmol in each ml of solution

Can be formulated :

\large \boxed {\bold {M ~ = ~ \dfrac {n} {V}}}

a.

V = 500 cm³ = 0.5 L

M = 2 mol/L

n=moles = M x V

n = 2 mol/L x 0.5 L

n = 1 mol

b.  

V = 2 L

M = 0.5 mol/L

n=moles = M x V

n = 0.5 mol/L x 2 L

n = 1 mol

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Explain where the weight of an atom is found what is responsible for that weight and why
padilas [110]

Answer:

The majority of the weight in an atom is found in the nucleus.

Explanation:

The protons and neutrons that make up the nucleus of the atom may take up a tiny amount of space in comparison to the rest of the atom, but they are far more dense than the electrons that orbit the nucleus.

6 0
3 years ago
A compound is found to contain 50. 05% sulfur and 49. 95% oxygen by mass. What is the empirical formula for this compound? SO S2
jekas [21]

The empirical formula for the given compound has been \rm SO_2. Thus, option C is correct.

The empirical formula has been the whole unit ratio of the elements in the formula unit.

<h3>Computation for the Empirical formula</h3>

The given mass of Sulfur has been, 50.05 g

The given mass of oxygen has been 49.95 g.

The moles of elements in the sample has been given by:

\rm Moles=\dfrac{Mass}{Molar\;mass}

  • Moles of Sulfur:

\rm Moles\;S=\dfrac{50.05}{32}\\&#10; Moles\;S=1.56\;mol

The moles of sulfur in the unit has been 1.56 mol.

  • Moles of Oxygen:

\rm Moles\;O=\dfrac{49.95}{16} \\&#10;Moles\;O=3.12\;mol

The moles of oxygen in the unit has been 3.12 mol.

The empirical formula unit has been given as:

\rm S_{1.56}O_{3.12}=SO_2

Thus, the empirical formula for the given compound has been \rm SO_2. Thus, option C is correct.

Learn more about empirical formula, here:

brainly.com/question/11588623

8 0
2 years ago
1. Do you think that the lightbulb and the Moon spheres are "to scale" compared to the real
Tems11 [23]

Answer:

Yes

Explanation: Had a question like this and I said yes and got it right

3 0
2 years ago
Calcuate the number of<br> grams of solute in 453.9mL<br> of 0.237 M calcium acetate
AysviL [449]

The number of  grams : 17.082 g

<h3>Further explanation</h3>

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

\large {\boxed {\bold {M ~ = ~ \dfrac {n} {V}}}

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

453.9 mL  of 0.237 M calcium acetate

  • mol

\tt mol=M\times V=0.237\times 0.4539=0.108

  • mass

MW Ca(C₂H₃OO)₂ : 158,17 g/mol

\tt mass=mol\times MW\\\\mass=0.108\times 158.17=17.082~g

4 0
3 years ago
POINTS FOR HELP!!!!!
Fudgin [204]
MgSO4, CO2, AlCl3 are the answer
3 0
3 years ago
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