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2 years ago

What element produces violet light of approximately 412 nm?

2 answers:

8 0

a. Hydrogen

Hydrogen has a line at 410 nm.
Mercury has a line at 405 nm.
Sodium and neon have no lines near 412 nm.

irinina [24]2 years ago

4 0

Answer:

a. Hydrogen

Explanation:

Emission spectrum of hydrogen produces four main visible wavelengths out of which one is violet light at 412 nm.

Mercury has visible prominent lines of blue, green and yellow-orange.

Sodium has two prominent very close yellow wavelengths in its emission spectrum.

Neon spectra also consists of green to red lines but not violet light.

Thus, correct option is a.

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A chlorine (CI) atom has 7 valence electrons. Which of the following would be the most likely way for a chlorine atom to become

pogonyaev

Answer:

Option C. Gain 1 electron

Explanation:

Valence electron(s) are the electron(s) located on the outermost shell of an atom. Valency is simply defined as the combining power of an atom.

Chlorine (Cl) atom has 7 valence electron. This implies that Cl needs just one electron to complete it's octet configuration. It will be difficult for Cl to lose any of it's valence electron(s). Cl can either gain or share 1 electron to become stable.

Thus, considering the options given in the question above, option C gives the correct answer to the question.

3 0

2 years ago

The solubility of lead(II) iodide is 0.064 g/100 mL at 20ºC. What is the solubility product for lead(II) iodide?

quester [9]

Answer:

$Ksp=1.07x10^{-8}$

Explanation:

Hello,

In this case, the dissociation reaction is:

$PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)$

For which the equilibrium expression is:

$Ksp=[Pb^{2+}][I^-]^2$

Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)

$Molar solubility=\frac{0.064g}{100mL}*\frac{1000mL}{1L}*\frac{1mol}{461.2g}=1.39x10^{-3}M$

In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:

$[Pb^{2+}]=1.39x10^{-3}M$

$[I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M$

Thereby, the solubility product results:

$Ksp=(1.39x10^{-3}M)(2.78x10^{-3}M)^2\\\\Ksp=1.07x10^{-8}$

Regards.

6 0

3 years ago

What is the rate of a reaction if the value of kis 0.1, [A] is 1 M, and [B] is 2 M?

mel-nik [20]

Answer:

D. 0.4 (mol/L)/S

Explanation:

You simply have to plug in the given values into the rate law.

Rate = k[A][B]

Rate = (0.1)(1)²(2)²

Rate = (0.1)(1)²(4)²

Rate = 0.4

8 0

3 years ago

What size volumetric flask would you use to create a 0.50 M solution using 10.00 g of CaBr2

SIZIF [17.4K]

Answer:

Explanation:

First convert the grams of Calcium Bromide to moles by using the atomic weight. Then use the formula for molarity, which is moles per liter.

CaBr2 = 199.9 g/mol

10/199.9 = 0.05 moles of CaBr2

$0.5M=\frac{0.05mol}{x}$

x = 0.1L or 100mL

5 0

2 years ago

A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. what is the partial pressure o

klasskru [66]

Hello!

We have the following statement data:

Data:
$P_{Total} = 800 mmHg$
$P\% N_{2} = 60\%$
$P\% O_{2} = 40\%$
$P_{partial} = ? (mmHg)$

As the percentage is the mole fraction multiplied by 100:

 $P = X_{ O_{2} }*100$

The mole fraction will be the percentage divided by 100, thus:
What is the partial pressure of oxygen in this mixture?

$X_{ O_{2} } = \frac{P}{100}$
$X_{ O_{2}} = \frac{40}{100}$
$\boxed{X_{ O_{2}} = 0.4}$

To calculate the partial pressure of the oxygen gas, it is enough to use the formula that involves the pressures (total and partial) and the fraction in quantity of matter:

In relation to $O_{2}$ :

$\frac{P O_{2} }{P_{total}} = X_O_{2}$
$\frac{P O_{2} }{800} = 0.4$
$P_O_{2} = 0.4*800$
$\boxed{\boxed{P_O_{2} = 320\:mmHg}}\end{array}}\qquad\quad\checkmark$

Answer:
b. 320.0 mm hg

7 0

3 years ago