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GenaCL600 [577]
2 years ago
12

Under what conditions does the hydrolysis of an amide bond occur?

Chemistry
1 answer:
Mademuasel [1]2 years ago
5 0
Hi.

I did some digging and I think I found what you're looking for.

I found this on Q(uizlet)

basic or acidic conditions and the reactants must be heated.

~
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Water is made of two [Blank] atoms and one [Blank] atom. (please fill in the blanks)​
natima [27]
Two [hydrogen] atoms and one [oxygen] atom
5 0
2 years ago
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How many atoms will there be in 5.00 g of gold. (Use ^ to show the exponent in scientific notation. For example, 4 X 1011 should
gayaneshka [121]

Answer:

1.53x10^22 atoms of Au

Explanation:

To find the atoms of gold we need first, to convert the mass of gold to moles using molar mass of gold (196.97g/mol). Then, these moles must be converted to number of atoms based on definition of moles (1 mole = 6.022x10²³ atoms).

<em>Moles Au:</em>

5.00g Au * (1mol / 196.97g) = 0.0254 moles of Au

<em>Atoms of Au:</em>

0.0254 moles * (6.022x10²³ atoms / 1 mole) =

<h3>1.53x10^22 atoms of Au</h3>
3 0
3 years ago
Explain why food chains do not tend to exceed four links.
Softa [21]
Energy is "lost" at each trophic level when you go up the chain. <span> Typically there are fewer organisms at higher trophic levels.

Hope this helps!</span>
3 0
2 years ago
The complete combustion of propane (C3H8) in the presence of oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) a
mixas84 [53]

Answer:

26.9 L is the volume of CO₂, we obtained

Explanation:

The reaction is: C₃H₈(g) + 5O₂(g)  →  3CO₂ (g) + 4H₂O (g)

Let's determine the reactants moles:

27.5 g . 1mol / 44 g = 0.625 moles

We need density of O₂ to determine mass and then, the moles.

O₂ density = O₂ mass / O₂ volume

O₂ density . O₂ volume = O₂ mass

1.429 g/L . 45L = O₂ mass → 64.3 g

Moles of O₂ → 64.3 g . 1mol/32g = 2.009 moles

Let's find out the limiting reactant:

1 mol of propane needs 5 moles of oxygen to react

Then, 0.625 moles will react with (0.625 . 5)/1 = 3.125 moles of O₂

Oxygen is the limiting reactant, we need 3.125 moles but we only have 2.009 moles

Ratio is 5:3. 5 moles of O₂ produce 3 moles of CO₂

Therefore, 2.009 moles of O₂ must produce (2.009 .3) /5 = 1.21 moles of CO₂. Let's find out the volume, by Ideal Gases Law (STP are 1 atm and 273K, the standard conditions)

1 atm . V = 1.21 moles . 0.082 . 273K

V = (1.21 moles . 0.082 . 273K) / 1atm = 26.9 L

3 0
3 years ago
How many potassium atoms are present in 0.01456 g of potassium??
leva [86]
Its going to be 2.81 x 1023 atoms 
8 0
2 years ago
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