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1 year ago

What does the arrangement of the periodic table help scientists see?

Chemistry

1 answer:

lina2011 [118]1 year ago

7 0

The relation between the elements electro negativity, atomic radius, and ionization energy levels.

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Is table salt an example of a substance?

bixtya [17]

Yes table salt is an example of a substance

7 0

1 year ago

Which one of the following statements is correct? View Available Hint(s) Which one of the following statements is correct? Adeni

Dmitrij [34]

Answer:

Adenine forms two hydrogen bonds with thymine;

guanine forms three hydrogen bonds with cytosine

Explanation:

There are three hydrogen bonds between guanine and cytosine. Between adenine and thymine, two hydrogen bonds form. Complementary pairs always include one base of purine and one base of pyrimidine.

4 0

1 year ago

Please can someone explain the answer to number 11

Aleks04 [339]

Answer:

HClO2 is the Lewis Acid and the ClO2- is the Lewis Base in the first pair

NH3 is the Lewis Base and the NH4+ is the Lewis Acid in the second pair.

Explanation:

A Lewis Base is an electron donor so there are extra pairs of electrons.

A Lewis Acid is an electron receiver so they are electron deficient.

7 0

1 year ago

423.3 mL of gas is at 49.2 C. It is compressed to a volume of 79 mL. What is the new temperture. Express your answer in Kelvin.

zzz [600]

Answer: 60.1K

Explanation:

Initial volume of gas V1 = 423.3mL

Initial temperature T1 = 49.2°C

Convert temperature in Celsius to Kelvin

( 49.2°C + 273 = 322.2K)

Final temperature T2 = ?

Final volume V2 = 79mL

According to Charle's law, the volume of a fixed mass of a gas is directly proportional to the temperature.

Mathematically, Charles' Law is expressed as: V1/T1 = V2/T2

423.3mL/322.2 = 79mL/T2

To get the value of T2, cross multiply

423.3mL x T2 = 322.2K x 79mL

423.3mL x T2 = 25453.8

T2 = (25453.8/423.3mL)

T2 = 60.1K

Thus, the new temperature of the gas is 60.1K

8 0

1 year ago

A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path

Igoryamba

Explanation:

For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

R = 8.314 J/(K mol), T = 298 K ,

$V_{2}$ = 8.34 L, $V_{1}$ = 2.67 L

Now, putting the given values into the above formula as follows.

W = $-2.303 nRT log(\frac{V_2}{V_1})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})$

= $-2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494$

= -2818.68 J

Hence, work for path A is -2818.68 J.

For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure $P_{external}$ = 1.00 atm.

So, W = $-P_{external} \times \Delta V$

Now, putting the given values into the above formula as follows.

W = $-P_{external} \times \Delta V$

= $-1 atm \times (8.34 L - 2.67 L)$

= -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

W = $-\frac{101.33 J}{1 atm L} \times 5.67 atm L$

= -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

W = -574.54 J

Hence, work for path B is -574.54 J.

3 0

1 year ago