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Sloan [31]
3 years ago
8

Element Z has a mass of 430g. Scientists determined it to have a ½ life of 5 years. How many grams of element Z will remain afte

r 15 years?

Chemistry
1 answer:
Afina-wow [57]3 years ago
7 0

Answer:

hope this helped youu:)

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3 years ago
Place the following compounds in order of increasing strength of intermolecular forces. Ch4 ch3ch2ch3 ch3ch3
Nikitich [7]

Explanation:

More is the mass of an alkane molecule more will be its boiling point. This means that forces of attraction will be more when mass is more as more heat is required to break the bonds within atoms. Hence, boiling point is more.

Therefore, in the given molecules mass of CH_{4} is 16.04 g/mol, mass of CH_{3}CH_{2}CH_{3} is 44 g/mol, and mass of CH_{3}CH_{3} is 30 g/mol.

Thus, we can see that CH_{3}CH_{2}CH_{3} has the highest mass and CH_{4} has least mass.

Hence, increasing strength of intermolecular forces of given compounds is as follows.

             CH_{4} < CH_{3}CH_{3} < CH_{3}CH_{2}CH_{3}          

8 0
3 years ago
Read 2 more answers
Alpha helices are a type of secondary structure in proteins. What is the length of a 33.0 kDa single‑stranded α‑helical protein
WARRIOR [948]

Answer:

450 Å

Explanation:

The amino acids in an α-helix are arranged in a right-handed helical structure where each amino acid residue corresponds to a 100° turn in the helix, and a translation of 1.5 Å (0.15 nm) along the helical axis

Mass of protein= 33.0 kDa

33.0 kDa ×(1000Da/kDa) × ( 1 residue/ 110 Da)

=33.0 ×1000 × 0.0091

= 300 residues

300 residues × (1.5Å /residue)

=450 Å

7 0
3 years ago
Which one of the following is the net ionic equation for the reaction of nitric acid with aluminum hydroxide
kompoz [17]
The ionic eqn is as follow:

1 Al(OH)3(s) + 3 H+(aq) + 3 NO3(-1) --> 1 Al(3+)(aq) + 3 NO3(-)(aq) + 3 H2O(l)

3moles of No3- ion on both sides cancels out to give the net ionic eqn:

1 Al(OH)3(s) + 3 H+(aq) --> 1 Al(3+)(aq) + 3 H2O(l)
5 0
3 years ago
A resting adult requires about 240 ml of pure oxygen/mind and breathes about 12 times every minute. if inhaled air contains 20 p
Dmitry_Shevchenko [17]

12 times breathe give 240 ml of pure O_{2}. Each breathe gives 20 ml of O_{2}.

Let us consider, volume of air per breathe= x ml.

Pure O_{2} from inhaled air= \frac{20}{100}x ml and Pure O_{2} from exhaled air= \frac{16}{100}x ml.

Pure O_{2} from inhaled and exhaled air= 20 ml

So, \frac{20}{100}x + \frac{16}{100}x = 20

Therefore, x = 55.5 ml

So, volume of air per breath= 55.5 ml.


7 0
3 years ago
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