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3 years ago

What is your guys ig?

Chemistry

2 answers:

makvit [3.9K]3 years ago

7 0

Answer:

dont have one

Explanation:

olga_2 [115]3 years ago

7 0

Answer:

ig means??

Explanation:

☜☆☞☜☆☞☜☆☞☜☆☞☜☆☞

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Cite two important quantum-mechanical concepts associated with the Bohr model of the atom.(b) Cite two important additional refi

Alenkinab [10]

Answer: Bohr postulated that electronic energy levels are quantized. Secondly, a photon of light of a particular frequency is emitted when electrons move from a higher to a lower energy levels.

Explanation:

The Bohr model of the atom is the immediate predecessor of the wave mechanical model of the atom. The wave mechanical model refined the Bohr's model by treating the electron as a wave having a wave function psi. The wave function describes the identity of the electron. From Heisenberg uncertainty principle, the position of a particle cannot be accurately and precisely measured. Hence the wave mechanical model added that electrons are not localized in orbits according to Bohr's model but the integral of psi squared dx gives the probability of finding the electron within a given space.

8 0

3 years ago

Read 2 more answers

What is the pH of a solution with a concentration of 1.8 × 10-4 molar H3O+?

Andre45 [30]

PH is defined as the negative log of Hydrogen ion concentration. Mathematically we can write this as:

$pH=-log[H^{+}]=-log[H_{3}O]$

We are given the concentration of $H_{3}O$ . Using the value in formula, we get:

$pH=-log[1.8*10^{-4}]=3.745$

Therefore, the pH of the solution will be 3.745

8 0

3 years ago

What are ionic compounds

ASHA 777 [7]

Answer: Ionic compounds are compounds consisting of ions.

Two-element compounds are usually ionic when one element is a metal and the other is a non-metal

Explanation: hope this helps!

7 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:

$\rm Moles\;S=\dfrac{50.05}{32}\\
 Moles\;S=1.56\;mol$

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

Moles of Oxygen:

$\rm Moles\;O=\dfrac{49.95}{16} \\
Moles\;O=3.12\;mol$

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

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2 years ago

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of ben

Nuetrik [128]

Here is the complete question.

Benzalkonium Chloride Solution ------------> 250ml

Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.

Sig: Dilute 10ml to a liter and apply to affected area twice daily

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?

(A) 1700 mL

(B) 29.4 mL

(D) 294 mL

Answer:

(B) 29.4 mL

Explanation:

1 L = 1000 mL

1:200 solution implies the $\frac{weight}{volume}$ in 200 mL solution.

200 mL of solution = 1g of Benzalkonium chloride

1000 mL will be $\frac{1000mL}{200mL}=\frac{1g}{xg}$

200mL × 1g = 1000 mL × x(g)

x(g) = $\frac{200mL*1g}{1000mL}$

x(g) = 0.2 g

That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.

∴ $\frac{10mL}{250mL}=\frac{0.2g}{y(g)}$

y(g) = $\frac{250mL*0.2g}{10mL}$

y(g) = 5g of benzalkonium chloride.

Now, at 17% $\frac{weight}{volume}$ concentrate contains 17g/100ml:

∴ the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;

= $\frac{17g}{5g} = \frac{100mL}{z(mL)}$

z(mL) = $\frac{100mL*5g}{17g}$

z(mL) = 29.41176 mL

≅ 29.4 mL

Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride

4 0

3 years ago