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zubka84 [21]
3 years ago
6

If you take a 15.0-mL portion of the stock solution and dilute it to a total volume of 0.600 L , what will be the concentration

of the final solution?
Chemistry
1 answer:
alexira [117]3 years ago
6 0

15 \div .6
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Which formula is an empirical formula?
poizon [28]
<h2>Hello!</h2>

The answer is:

The empirical formula is the option B. NH_{3}

<h2>Why?</h2>

The empirical formula of a compound is the simplest formula that can be written. On the opposite, the molecular formula involves a variant of the same compound, but it can be also simplified to an empirical formula.

MolecularFormula=n(EmpiricalFormula)

We are looking for a formula that cannot be simplified by dividing the number of molecules/atoms that conforms the compound.

Let's discard option by option in order to find which formula is an empirical formula (cannot be simplified)

A. N_{2}O_{4}

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

N_{2}O_{4}=2(NO_{2})

B. NH_{3}

It's an empirical formula since it cannot be obtained by the multiplication of a whole number and the simplest formula. It's the simplest formula that we can find of the compound.

C. C_{3}H_{6}

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

C_{3}H_{6}=3(CH_{2})

D. P_{4}O_{10}

It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

P_{4}O_{10}=2(P_{2}O_{5})

Hence, the empirical formula is the option B. NH_{3}

Have a nice day!

6 0
3 years ago
Read 2 more answers
The following equation is one way to prepare oxygen in a lab. 2KClO3 → 2KCl + 3O2 Molar mass Info: MM O2 = 32 g/mol MM KCl = 74
Alex73 [517]
From  the  equation  above   the  reacting   ratio  of  KClO3   to  O2  is  2:3 therefore  the  number  of  moles  of  oxygen  produced  is  ( 4 x3)/2 =  6 moles  since   four  moles  of  KClO3  was  consumed
mass=relative  formula mass  x  number  of  moles
That  is   32g/mol x 6  moles  =192grams


8 0
3 years ago
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Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimen
olga55 [171]

Answer: Rate law=k[A]^1[B]^2, order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is 3L^2mol^{-2}s^{-1}

Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

Rate=k[A]^x[B]^y

k= rate constant

x = order with respect to A

y = order with respect to A

n = x+y = Total order

a) From trial 1: 1.2\times 10^{-2}=k[0.10]^x[0.20]^y    (1)

From trial 2: 4.8\times 10^{-2}=k[0.10]^x[0.40]^y    (2)

Dividing 2 by 1 :\frac{4.8\times 10^{-2}}{1.2\times 10^{-2}}=\frac{k[0.10]^x[0.40]^y}{k[0.10]^x[0.20]^y}

4=2^y,2^2=2^y therefore y=2.

b) From trial 2: 4.8\times 10^{-2}=k[0.10]^x[0.40]^y    (3)

From trial 3: 9.6\times 10^{-2}=k[0.20]^x[0.40]^y   (4)

Dividing 4 by 3:\frac{9.6\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.20]^x[0.40]^y}{k[0.10]^x[0.40]^y}

2=2^x,2=2^1, x=1

Thus rate law is Rate=k[A]^1[B]^2

Thus order with respect to A is 1 , order with respect to B is 2 and total order is 1+2=3.

c) For calculating k:

Using trial 1:  1.2\times 10^{-2}=k[0.10]^1[0.20]^2

k=3 L^2mol^{-2}s^{-1}.



6 0
3 years ago
2.What two factors must be held constant for density to be a constant ratio?
irina1246 [14]

Answer:

Temperature and Pressure

Explanation:

Temperature and pressure cause change in volume.

So any change in volume will alter the ratio of density as given by equation of density.

Density = mass/ volume

Change in volume will alter the ratio.

Kindly mark it branliest if the answer is little bit satisfying.

6 0
3 years ago
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How spectral lines are formed
Ket [755]

=Spectral lines are produced by transitions of electrons within atoms or ions. As the electrons move closer to or farther from the nucleus of an atom (or of an ion), energy in the form of light (or other radiation) is emitted or absorbed.…

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3 years ago
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