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riadik2000 [5.3K]
2 years ago
10

HURRY ! Kumar is producing the photoelectric effect by using red light. He wants to increase the energy of emitted electrons. Ba

sed on the research of Albert Einstein, what is the best way for him to do this? Increase the intensity of red light. Decrease the intensity of red light. Use a different colored light that has a higher frequency. Use a different colored light that has a lower frequency.
Chemistry
1 answer:
RideAnS [48]2 years ago
4 0

Answer:

using a diffrent colored light with a higher frequency

Explanation:

You might be interested in
Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K
Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

And the undergoing chemical reaction:

MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0
3 years ago
Need help answering
svp [43]

Answer:

Your answer would be E

Explanation:

4 0
2 years ago
How many moles of KBr are present in 500 ml of a 0.8 M KBr solution?
faltersainse [42]

Answer:

2) 0.4 mol

Explanation:

Step 1: Given data

  • Volume of the solution (V): 500 mL
  • Molar concentration of the solution (M): 0.8 M = 0.8 mol/L

Step 2: Convert "V" to L

We will use the conversion factor 1 L = 1000 mL.

500 mL × 1 L/1000 mL = 0.500 L

Step 3: Calculate the moles of KBr (solute)

The molarity is the quotient between the moles of solute (n) and the liters of solution.

M = n/V

n = M × V

n = 0.8 mol/L × 0.500 L = 0.4 mol

4 0
3 years ago
What is the mass of 7.50 moles of magnesium chloride, mgcl2?
Natali5045456 [20]
MgCl2 = 1Mg + 2Cl = 1(24.3) + 2(35.45) = 95.2g/1mole
7.50moles MgCl2 x 95.2g MgCl2 = 714g MgCl2
8 0
3 years ago
Which ion is the only negative ion produced by an Arrhenius base in water?
Snowcat [4.5K]
According to Arrhenius Theory of acids and bases, an Arrhenius base, when dissolving in water, produces the only negative ion: OH-. 

Therefore, (3) OH- is the correct answer.

Hope this is helpful~
6 0
3 years ago
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