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3 years ago

How many distinct tripeptides can be made from the two amino acids leucine and histidine?

Chemistry

1 answer:

ValentinkaMS [17]3 years ago

5 0

The right answer is 8 (2^3)

to solve these kind of question, you should use this equation:

solution = number of kinds of elements (aminoacids) ^ the lenght of the sequence (polypeptide).

If they are not numerous you can write those possibilities: (K for lysine and H for histidine).

1. KKK

2. KKH

3. KHK

4. KHH

5. HKK

6. HKH

7. HHK

8. HHH

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Which of the following is the best name for a compound made from calcium and bromine? (CaBr2) calcium bromide calcium dibromide

Mice21 [21]

Answer:

Calcium bromide

Explanation:

When naming compounds, the use of prefixes depend on the type of bond made. In this case, calcium and bromine form a ionic bond because calcium is a metal and bromine is a non-metal.

Ionic bonds are not named using prefixes. So no matter how many atoms there are, you will simply write the name of the element for the first element.

For the second element, you name it as well, but only use the root name and end it with -ide.

7 0

4 years ago

Total number of atoms in 4so2

Sunny_sXe [5.5K]

Answer:

s=8

o=2

Hope it helps to

3 0

3 years ago

Write the chemical formula of tetraphosphours octasulfide

ArbitrLikvidat [17]

Answer:

P4 S8

Explanation:

8 0

3 years ago

TRUE or FALSE Nonpolar molecules only have nonpolar bonds.

Kitty [74]

Answer:

its false

Explanation:

srry if its wrong ;-;

4 0

3 years ago

Bromine (63 g) and fluorine (60 g) are mixed to give bromine trifluoride. a) Write a balanced chemical reaction. b) What is the

fiasKO [112]

Answer:

a) Br2 + 3F2 → 2BrF3

b) Br2 is the limiting reactant.

c) There will be formed 86.3 grams of BrF3

d) There will remain 0.326 moles of F2 = 12.97 grams

Explanation:

Step 1: The balanced equation

Br2 + 3F2 → 2BrF3

Step 2: Given data

Mass of Bromine = 63grams

Mass of fluorine = 60 grams

percent yield = 80%

Molar mass of bromine = 79.9 g/mol =

Molar mass of fluorine = 19 g/mol

Molar mass of bromine trifluoride = 136.9 g/mol

Step 3: Calculating moles

Moles Br2 = 63 grams / (2*79.9)

Moles Br2 = 0.394 moles

Moles F2 = 60 grams / (2*19.9)

Moles F2 = 1.508

For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

Br2 is the limiting reactant. It will completely be consumed (0394 moles).

There will react 3*0.394 = 1.182 moles of F2

There will remain 1.508 - 1.182 = 0.326 moles of F2 = 12.97 grams

Step 5: Calculate moles of BrF3

For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

So there is 2*0.394 moles = 0.788 moles of BrF3 moles produced

Step 6: Calculate mass of BrF3

mass = Moles * Molar mass

mass of BrF3 = 0.788 moles * 136.9 g/mol = 107.88 grams = Theoretical yield

Step 7: Calculate actual yield

% yield = 0.80 = actual yield / theoretical yield

actual yield = 0.80 * 107.88 grams = 86.3 grams

actual yield = 86.3 grams

7 0

3 years ago