Answer:
Adding more substrate would overcome the effect of the compound
Explanation:
- Enzymes are biochemical catalysts that speed up chemical reactions. They act on specific substrate to convert them to products.
- Compounds known as inhibitors slow down the rate of enzyme activity.
- Inhibitors are classified as competitive and non-competitive inhibitors.
- Competitive inhibitors will compete with the substrate to bind the active sites on the enzyme. The effect of competitive inhibitors may be reduced by increasing the concentration of the substrate.
- The compound added by the biologist was a competitive inhibitor and therefore adding more substrate would overcome its effect on enzyme catalysis
- Non-competitive inhibitors binds the active site of the enzyme permanently and prevents the substrate from accessing the active sites.
Answer:
975.56×10²³ molecules
Explanation:
Given data:
Number of molecules of C₂H₆ = 4.88×10²⁵
Number of molecules of CO₂ produced = ?
Solution:
Chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Number of moles of C₂H₆:
1 mole = 6.022×10²³ molecules
4.88×10²⁵ molecules×1mol/6.022×10²³ molecules
0.81×10² mol
81 mol
Now we will compare the moles of C₂H₆ with CO₂.
C₂H₆ : CO₂
2 : 4
81 : 4/2×81 = 162 mol
Number of molecules of CO₂:
1 mole = 6.022×10²³ molecules
162 mol ×6.022×10²³ molecules / 1 mol
975.56×10²³ molecules
The molecules or atoms that are formed by gain or loss of one or more valence electrons are said to be ions.
When atom loss one or more valence electrons, results in formation of cation whereas when atom gain one or more valence electrons, then formation of anion occurs. Cations carry positive charge and anions carry negative charge.
In general, cations are smaller than the neutral atoms from which they are formed and anions are larger than the neutral atoms.
As cations are smaller than the related neutral atoms because the valence electrons are lost which are farthest away from the nucleus. After that, taking more electrons distant from the cation results in reduction of radius of the ion.
Thus, aluminium cation consist of few electrons which results in fewer occupied energy levels by the electrons further results in reduction of radius i.e. smaller size.
Hence, given statement is true i.e. aluminium atom is larger than the aluminium cation as cation has fewer occupied energy levels.
We have to first write a balanced equation.
so2 + o2 -> so3
this is not balanced though. we have 3 oxygen on right and 4 on left
2so2 + o2 -> 2so3
now it is same on both sides. we have to figure out which is limiting reagent with the given amounts of reagents. we do this by comparing the ratio between them in terms of moles. we see that so2 has a coefficient of 2 and o2 has none which implies 1 and so3 has 2. this means that for every 2 moles of so2 reacting with 1 mole of o2, we get 2 moles of so3.
lets convert the given values to moles. to do this we know that molecular weight is measured in grams per mole. we are given grams and need to cancel out the grams to get moles. so the molecular weight:
so2 =32.1 + 2 * 16 = 64.1 g/mol
o2 = 2 * 16 = 32 g/mol
so3 = 32.1 + 3 * 16 = 80.1 g/mol
now to convert 90 g of 2so2 under ideal conditions.
90g / 64.1g/mol = 1.404 moles
convert this amount of moles of so2 to moles of o2. we have 2 moles of so2 to 1 of o2
1.404moles so2 / 2 moles so2 * 1 mole o2= 0.702 moles o2
so we see under ideal conditions that 90g of so2 would react with .702g of o2. lets see how many we actually have with 100g of o2
100g / 32g/mol =3.16 mol.
so we have a lot more o2 than needed. we are looking for how much is left in grams. we have to figure out how much was used. to do this convert our ideal moles of o2 into grams.
.702 moles o2 * 32g/mol = 22.5g o2
so what we startrd with (100g) minus what we needed (22.5g) is what we have left
100 - 22.5 = 77.5g o2
