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:
The metal has a heat capacity of 0.385 J/g°C
This metal is copper.
Explanation:
<u>Step 1</u>: Data given
Mass of the metal = 21 grams
Volume of water = 100 mL
⇒ mass of water = density * volume = 1g/mL * 100 mL = 100 grams
Initial temperature of metal = 122.5 °C
Initial temperature of water = 17°C
Final temperature of water and the metal = 19 °C
Heat capacity of water = 4.184 J/g°C
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<u>Step 2: </u>Calculate the specific heat capacity
Heat lost by the metal = heat won by water
Qmetal = -Qwater
Q = m*c*ΔT
m(metal) * c(metal) * ΔT(metal) = - m(water) * c(water) * ΔT(water)
21 grams * c(metal) *(19-122.5) = -100 * 4.184 * (19-17)
-2173.5 *c(metal) = -836.8
c(metal) = 0.385 J/g°C
The metal has a heat capacity of 0.385 J/g°C
This metal is copper.
<span>The relative strength of intermolecular forces such as ionic, hydrogen bonding, dipole-dipole interaction and Vander Waals dispersion force affects the boiling point of a compound. For this case, the longer the chain the higher the boiling point.
</span>CH, CH4, C4H10, C8H18, C16H34
Hope this answers the question. Have a nice day.
Answer:
evapouration is a type of vaporization that occur on the surface of liquid as it change into the gas phase.
hope it helps.
Answer:
See below, please
Explanation:
1 mol CH4 ——＞2 mol H2O=2x18 g/mol= 36 g
The question is not clear. You should tell us the concentration of CH4, methane.
