I don't know if this is the right subject but PLS HELP

Which of the following is true of enzymes? Enzymes may require a nonprotein cofactor or ion for catalysis to take speed up more

Sergio [31]

Answer:

Enzymes may require a nonprotein cofactor or ion for catalysis to take speed up more appreciably than if the enzymes act alone;

Enzymes increase the rate of chemical reaction by lowering activation energy barriers.

Explanation:

Some enzymes need a cofactor to act, it is attached to the enzyme and can be nonprotein such as a metal ion. The enzyme function depends on the physical properties of the environmental, especially temperature and pH, each enzyme has a great point of pH and temperature where it has a maximum activity.

If the three-dimensional function of an enzyme is altered, it loses it specified and may not catalyze the reaction, because the structure of the enzyme is responsable for its specified. The catalyst occurs because the enzyme lows the activation energy barriers and this increases the rate of the reaction.

7 0

3 years ago

Identify the variables in this hypothesis. If the pressure of a gas is increased, then the volume will decrease because the part

Kaylis [27]

Volume of matter decreases under pressure ... -under pressure, the particles in a gas are forced closer together ... factors affecting gas pressure ... -if pressure in a sealed container is lower than outside, gas will rush in ...

7 0

4 years ago

Read 2 more answers

I need to know the answer to that

Tresset [83]

im not sure which one to answer, and i can hardly see the text.

8 0

3 years ago

What is the MOLAR heat of combustion of methane(CH₄) if 64.00g of methane are burned to heat 75.0 ml of water from 25.00°C to 95

melamori03 [73]

Answer:

-5.51 kJ/mol

Explanation:

Step 1: Calculate the heat required to heat the water.

We use the following expression.

$Q = c \times m \times \Delta T$

where,

c: specific heat capacity
m: mass
ΔT: change in the temperature

The average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.

$Q = 4.184J/g.\°C \times 75.0g \times (95.00\°C - 25.00\°C) = 2.20 \times 10^{3} J = 2.20 kJ$

Step 2: Calculate the heat released by the methane

According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero

Qc + Qw = 0

Qc = -Qw = -22.0 kJ

Step 3: Calculate the molar heat of combustion of methane.

The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.

$\frac{-22.0kJ}{64.00g} \times \frac{16.04g}{mol} = -5.51 kJ/mol$

8 0

3 years ago

The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A

liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

$p^{OH}$ =14-56.17

=8.823

The volume of the $NH_{3}$ = 40.00 ml

convert into the liter= 0.040L

The value of the concentrated $NH_{3}$ =0.10 M

The volume of the $NH_{4}Cl$ = 50.00 ml

convert into the liter= 0.050L

The value of concentrated $NH_{4}Cl$ = 0.10 M

The volume of the $H_{2}So_{4}$ = 30 ml

convert into the liter= 0.030L

The value of concentrated $H_2So_4$ =0.05 M

Calculating total volume=(0.40+0.050+0.030)

=0.120 L

calculating the new concentrated value of $NH_3$ = $\frac{0.10\times 0.040}{0.120}= 0.33 \ M$

calculating the new concentrated value of $NH_4Cl$ = $\frac{0.050\times 0.10}{0.120}= 0.04166 \ M$ calculating the new concentrated value of $H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M$ when 1 mol $H_2So_4$ produced 2 mols $H^{+}$ so, 0.0125 in $H_2So_4$ produced:

$=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}$

create the ICE table:

$NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}$

I (m) 0.033(m) 0.025 0.04166

C -0.025 -0.025 + 0.025

E 8.3\times 10^{-3} 0 0.0667

now calculating pH:

when ph= 8.83:

$P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769$

5 0

3 years ago