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ASHA 777 [7]
2 years ago
9

based on student 1’s explanation, the same substance composes both of the samples in which of the following pairs?

Chemistry
1 answer:
Nikitich [7]2 years ago
8 0
What was student 1 ???
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Why do atoms share electrons in covalent bonds??
ra1l [238]
Atoms in covalent bonds do combine so as to be stable. As covalent bond consist non metals e.g O2 in this example each atom has vacance of 2 orbitals/ electrons so shairing electrons result their stability
6 0
3 years ago
Read 2 more answers
What is the standard enthalphy change ΔHo, for the reaction represented above? (ΔHof of C2H2(g) is 230 kJ mol-1; (ΔHof of C6H6(g
wolverine [178]

Answer:

-608KJ/mol

Explanation:

3 C2H2(g) -> C6H6(g)

ΔHrxn = ΔHproduct - ΔHreactant

ΔHrxn= ΔHC6H6 - 3ΔHC2H2

ΔHrxn = 83 - 3(230)

ΔHrxn = -608

6 0
3 years ago
Which one is NOT part of the cell theory?
worty [1.4K]

<u>Answer:</u>

All living things are not made of cells.

5 0
2 years ago
how many grams of antifreeze would be required per 500 g of water to prevent the water from feezing at a temperature of -39° C​
andrezito [222]

Answer:

333.7g of antifreeze

Explanation:

Freezing point depression in a solvent (In this case, water) occurs by the addition of a solute. The law is:

ΔT = Kf × m × i

Where:

ΔT is change in temperature (0°C - -20°C = 20°C)

Kf is freezing point depression constant (1.86°C / m)

m is molality of solution (moles solute / 0.5 kg solvent -500g water-)

i is Van't Hoff factor (1, assuming antifreeze is ethylene glycol -C₂H₄(OH)₂)

Replacing:

20°C = 1.86°C / m  × moles solute / 0.5 kg solvent × 1

5.376 = moles solute

As molar mass of ethylene glycol is 62.07g/mol:

5.376 moles × (62.07g / 1mol) = <em>333.7g of antifreeze</em>.

4 0
3 years ago
The following combinations are not allowed. If n and ml are correct, change the l value to create an allowable combination:
motikmotik

The allowable combination for the atomic orbital is n=3, l=1, m_{l}=-1.

<h3>What are the three quantum numbers of an atomic orbital?</h3>

Three quantum numbers specify an atomic orbital:

- The principal quantum number, n, which is a positive integer, describes the relative size of the orbital and its distance from the nucleus.

- l is the angular momentum quantum number that is related to the shape of the orbital; l is an integer from 0 to n-1 (so n limits l ),

- $\boldsymbol{m}_{l}$ is the magnetic quantum number that prescribes the three-dimensional shape of the orbital around the nucleus; m_{l} values are integers from -l to =l(l limits ml)

For n = 3, l can have three values: 0, 1, and 2. Since m_{l} values are integers from -l to 0 to +l, for l = 0 the value of m_{l} cannot be -1 (l = 0 has m_{l}= 0).

There are two l values that are consistent with n and m_{l} values:

l=1 or 2

Therefore, the allowable condition is n=3, l=1, m_{l}=-1.

To know more about quantum numbers, visit: brainly.com/question/16979660

#SPJ4

4 0
1 year ago
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