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mixas84 [53]
3 years ago
8

Light of a certain energy shines on a metal and causes electrons to be emitted.

Chemistry
2 answers:
sashaice [31]3 years ago
8 0

your answer is: the use of that light has a lower frequency

Inessa [10]3 years ago
6 0

Light of certain energy shines on a metal and causes electrons to be emitted. Based on the research of Albert Einstein, the change that would most likely result in stopping the emission of electrons from this metal is to coat the metal.

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Which of the following numerical expressions gives the number of particles in 2.0 g of Ne?
mojhsa [17]

Answer:

Number of particles = 2.0 g*(6.0 x 10^23 particles/mol) / 20.18 g/mol

Option C is correct

Explanation:

Step 1: Data give

Mass of Ne = 2.0 grams

Molar mass of neon = 20.18 g/mol

Number of Avogadro = 6.0 *10^23 /mol

Step 2: Calculate number of moles of neon

Moles Ne = Mass of ne / Molar mass of ne

Moles Ne = 2.0 / 20.18 g/mol

Moles Ne = 0.099 moles

Step 3: Calculate nulber of particles

Number of particles = 6.022*10^23 / mol * 0.099 moles = 5.96 *10^22

Number of particles = 6.022*10^23 * (2.0g/ 20.18g/mol)

Number of particles = 2.0 g*(6.0 x 10^23 particles/mol) / 20.18 g/mol

Option C is correct

7 0
3 years ago
Which of the following actions will not increase the rate at which a solid dissolves in
aliya0001 [1]

Answer:

O lowering the temperature of the system

5 0
3 years ago
A chunk of copper has a volume of 1.45 mL and a mass of 12.93 grams. What is the density of the chunk of copper?
natta225 [31]

Answer:

Density = 8.92 \ g/mL

Explanation:

The equation for density is:

Density = \frac{mass}{volume}

We plug in the given values:

Density = \frac{12.93 \ g}{1.45 \ mL}

Density = 8.92 \ g/mL

7 0
3 years ago
Read 2 more answers
Which of the following is considered a renewable resource?
Mashutka [201]

Explanation:

I can give you some examples;

1) water

2) biomass

3)Soil

4) forest...

I hope this will help you

5 0
3 years ago
Read 2 more answers
An unknown compound contains 38.7 % calcium, 19.9 % phosphorus and 41.2 % oxygen. what is the empirical formula of this compound
Temka [501]

The amount per 100 g is:

38.7 % calcium = 38.7g Ca / 100g compound = 38.7g

19.9 % phosphorus = 19.9g P / 100g compound = 19.9g

41.2 % oxygen = 41.2g O / 100g compound = 41.2g

The molar amounts of calcium, phosphorus and oxygen in 100g sample are calculated by dividing each element’s mass by its molar mass:

Ca = 38.7/40.078 = 0.96

P = 19.9/30.97 = 0.64

O = 41.2/15.99 = 2.57

C0efficients for the tentative empirical formula are derived by dividing each molar amount by the lesser value that is 0.64 and in this case, after that multiply wih 2.

Ca = 0.96 / 0.64 = 1.5=1.5 x 2 = 3

P = 0.64 / 0.64 = 1 = 1x2= 2

O = 2.57 / 0.64 = 4= 4x2= 8

Since, the resulting ratio is calcium 3, phosphorus 2 and oxygen 8

<span>So, the empirical formula of the compound is Ca</span>₃(PO₄)₂

5 0
3 years ago
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