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DochEvi [55]
3 years ago
6

RtR Naming Test Review Name: Per: Name the non-metal elements

Chemistry
1 answer:
Over [174]3 years ago
7 0
Non-Metals:
Hydrogen
Oxygen
Carbon
Nitrogen
Fluorine
Phosphorus
Sulfur
Chlorine
Selenium
Bromine
Iodine
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I think No

Explanation:

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Nuclear Processes Unit Test
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Select the options that are properties of electromagnetic waves:
kaheart [24]

Answer:

They propagate in materialistic media and non-materialistic media ( space ) .

They propagate in the space at constant velocity , which is 3 × 108 m/s .

They consist of vibrating electric and magnetic fields at certain frequency in phase with each other , perpendicular to each other and perpendicular to the direction of the wave propagation ...

Explanation:

welcome!

4 0
3 years ago
The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be
Mrac [35]

<u>Answer:</u> The freezing point of solution is 5.35°C

<u>Explanation:</u>

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 4.90°C/m

m_{solute} = Given mass of solute (naphthalene) = 2.60 g

M_{solute} = Molar mass of solute (naphthalene) = 128.2 g/mol

W_{solvent} = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC

Hence, the freezing point of solution is 5.35°C

3 0
3 years ago
In a reaction where an OH group is substituted with a Cl, 15.02 g of 3-phenyl-2-butanol (MW=150.2 g/mol) is the starting materia
Brrunno [24]

Answer:

16.81 g of C₁₀H₁₃Cl

Explanation:

When 3-phenyl-2-butanol reacts with HCl it undergoes a substitution reaction to form 2-chloro-3-phenylbutane and water.

C₁₀H₁₄O + HCl ⇄ C₁₀H₁₃Cl + H₂O

The molar mass of 3-phenyl-2-butanol is 150.2 g/mol and the molar mass of 2-chloro-3-phenylbutane is 168.1 g/mol. The molar ratio is 1:1. Then, for 15.02 g of 3-phenyl-2-butanol, the theoretical yield is:

15.02gC_{10}H_{14}O.\frac{168.1gC_{10}H_{13}Cl}{150.2gC_{10}H_{14}O} =16.81gC_{10}H_{13}Cl

8 0
4 years ago
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