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il63 [147K]
4 years ago
7

Which of the following invisible marks represents an inserted tab

Chemistry
1 answer:
Andrej [43]4 years ago
8 0
There are no invisible marks
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Classify each of the following as an element a compound or a mixture ​
Soloha48 [4]
Is there something else that goes along with the question?
6 0
3 years ago
When you add an inert gas, the reaction
julsineya [31]

Answer: When an inert gas is added to the system in equilibrium at constant pressure, then the total volume will increase.

Explanation:

3 0
3 years ago
There are over 100 elements, but only
elena-14-01-66 [18.8K]

Answer:

Symbol_Element

He_Helium

Li_Lithium

Be_Beryllium

B_Boron

C_Carbon

N_Nitrogen

O_Oxygen

F_Fluorine

Ne_Neon

Na_Sodium

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Al_Aluminum

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P_Phosphorous

S_Sulfur

Cl_Chlorine

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6 0
4 years ago
How many times higher is the concentration of H+ in the Hubbard Brook sample than in unpolluted rainwater?
Anna [14]

Answer:

1. 7 (a neutral solution)

Answer: 10-7= 0.0000001 moles per liter

2. 5.6 (unpolluted rainwater)

Answer: 10-5.6 = 0.0000025 moles per liter

3. 3.7 (first acid rain sample in North America)

Answer: 10-3.7 = 0.00020 moles per liter

The concentration of H+ in the Hubbard Brook sample is 0.00020/0.0000025, which is 80 times higher than the H+ concentration in unpolluted rainwater.

Explanation:

8 0
3 years ago
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
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