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4 years ago

PLEASE HELP ME!!! SCIENCE QUESTION!!!

Chemistry

1 answer:

sammy [17]4 years ago

8 0

Your answer is b main sequence

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2. Calculate the concentration (in molarity) of a NaOH solution if 25.0 mL of the solution are needed to neutralize 17.4 mL of a

zimovet [89]

Answer:

0.217 M NaOH

Explanation:

M1V1 = M2V2

M1 = 0.312 M HCl

V1 = 17.4 mL HCl

M2 = ?

V2 = 25.0 mL NaOH

Solve for M2 --> M2 = M1V1/V2

M2 = (0.312 M)(17.4 mL) / (25.0 mL) = 0.217 M NaOH

3 0

3 years ago

The electron configuration belonging to the atom with the highest second ionization energy is ________.

Gemiola [76]

Answer:

Element Lithium

Explanation:

The element with the highest second ionization energy is lithium. It belongs to the alkaline metal group I.e group one metals

It has the highest second ionization energy because it is very difficult to remove the electron from the 1s orbital.

Its atomic number is 3. The electronic configuration is 1s2 2S1

5 0

3 years ago

URGENT PLZ HELP

kramer

Answer:

Explanation:

In an endothermic reaction heat is applied.

4 0

3 years ago

Hydrogen sulfide (H2S) is a gas that smells like rotten eggs. It can be produced by bacteria in your mouth and contributes to ba

Doss [256]

Answer:

See explanation.

Explanation:

Hello!

In this case, we consider the questions:

a. Ideal gas at:

i. 273.15 K and 22.414 L.

ii. 500 K and 100 cm³.

b. Van der Waals gas at:

i. 273.15 K and 22.414 L.

ii. 500 K and 100 cm³.

Thus, we define the ideal gas equation and the van der Waals one as shown below:

$P^{id}=\frac{nRT}{V}\\\\ P^{vdW}=\frac{RT}{v-b}-\frac{a}{v^2}$

Whereas b and a for hydrogen sulfide are 0.0434 L/mol and 4.484 L²*atm / mol² respectively, therefore, we proceed as follows:

i. 273.15 K and 22.414 L.

$P^{id}=\frac{1.00mol*0.08206\frac{atm*L}{mol*K}*273.15K}{22.414L}=1 .00 atm$

ii. 500 K and 100 cm³ (0.1 L).

$P^{id}=\frac{1.00mol*0.08206\frac{atm*L}{mol*K}*500K}{0.100L}=410.3 atm$

i. 273.15 K and 22.414 L: in this case, v = 22.414 L / 1.00 mol = 22.414 L/mol

$P^{vdW}=\frac{0.08206\frac{atm*L}{mol*K}*273.15K}{22.414L/mol-0.0434L/mol}-\frac{4.484 atm*L^2/mol^2}{(22.414L/mol)^2}=0.993atm$

ii. 500 K and 100 cm³: in this case, v = 0.1 L / 1.00 mol = 0.100 L/mol

$P^{vdW}=\frac{0.08206\frac{atm*L}{mol*K}*500K}{0.100L/mol-0.0434L/mol}-\frac{4.484 atm*L^2/mol^2}{(0.100L/mol)^2}=276.5atm$

Whereas we can see a significant difference when the gas is at 500 K and occupy a volume of 0.100 L.

Best regards!

5 0

3 years ago

The use of light energy to convert water and carbon dioxide into energy-rich glucose molecules is called ____?

Lynna [10]

Answer isphotosynthesis

4 0

4 years ago

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