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3 years ago

For a specific enzyme/substrate system, which condition will always result in an increase in Vo (assuming nothing else changes)?

1 answer:

8 0

<span>a decrease in Km and an increase in Vmax
km is the constant that relates how thick the volume of the substrate in terms of concentration when the speed is a large portion of the most extreme. They increment Km by meddling with the joining process that makes them stick together wit the substrate, however they have any effect with regards to Vmax since since the sticking process to ES did not occur</span>

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Suppose that 0.48 g of water at 25 ∘ C condenses on the surface of a 55- g block of aluminum that is initially at 25 ∘ C . If th

mamaluj [8]

Answer:

$49^oC$

Explanation:

At $25^oC$ , the heat of vaporization of water is given by:

$\Delta H^o_{vap} = 43988 J/mol\cdot \frac{1 mol}{18.016 g} = 2441.6 J/g$

The water here condenses and gives off heat given by the product between its mass and the heat of vaporization:

$Q_1 = \Delta H^o_{vap} m_w$

The block of aluminum absorbs heat given by the product of its specific heat capacity, mass and the change in temperature:

$Q_2 = c_{Al}m_{Al}(t_f - t_i)$

According to the law of energy conservation, the heat lost is equal to the heat gained:

$Q_1 = Q_2$ or:

$\Delta H^o_{vap} m_w = c_{Al}m_{Al}(t_f - t_i)$

Rearrange for the final temperature:

$\Delta H^o_{vap} m_w = c_{Al}m_{Al}t_f - c_{Al}m_{Al}t_i$

We obtain:

$\Delta H^o_{vap} m_w + c_{Al}m_{Al}t_i = c_{Al}m_{Al}t_f$

Then:

$t_f = \frac{\Delta H^o_{vap} m_w + c_{Al}m_{Al}t_i}{c_{Al}m_{Al}} = \frac{2441.6 J/g\cdot 0.48 g + 0.903 \frac{J}{g^oC}\cdot 55 g\cdot 25^oC}{0.903 \frac{J}{g^oC}\cdot 55 g} = 49^oC$

6 0

3 years ago

The reducing agent in this catalytic hydrogenation reaction was molecular hydrogen (H2), which was produced in situ (in the reac

creativ13 [48]

Answer:

a) After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) $4NaBH_{4} + 2HCl + 7H_{2}O$ ⇒ 16 $H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}$ c) No $H_{2}$ was not the limiting reactant.

Explanation:

Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.

a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.

b) $4NaBH_{4} + 2HCl + 7H_{2}O$ ⇒ 16 $H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}$

c) $H_{2}$ was not the limiting reactant based on the mol to mol ratio of $H_{2}$ and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of $H_{2}$ would also be produced.

4 0

3 years ago

According to Table I, which equation represents a change resulting in the greatest quantity of energy released?

love history [14]

The answer is 3. The releasing of energy means exothermic reaction. So the ΔH should be negative. And the greatest quantity of energy released means that the greatest number. So according to the table I, the answer is 3.

8 0

3 years ago

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When you breathe the first step is that your lungs expand. why is this step needed

Alinara [238K]

Answer:

When you breathe in, or inhale, your diaphragm contracts and moves downward. This increases the space in your chest cavity, and your lungs expand into it. The muscles between your ribs also help enlarge the chest cavity. They contract to pull your rib cage both upward and outward when you inhale.

Explanation:

^^^

5 0

3 years ago

A stock solution has a concentration of 1.5 M NaCl and is diluted to a 0.80 M solution with a volume of 0.10 L. What volume of t

ioda

Answer:

0.053 L is the volume of concentrated solution that was used

Explanation:

Let's determine the answer of this, by rules of three.

There is also a dilution formula.

Molarity is a sort of concentration that indicates the moles of solute in 1L of solution.

In 1 L of concentrated solution, there are 1.5 moles of NaCl

In 1 L of diluted solution, there are 0.80 moles.

The volume for the diluted solution is 0.10L

The rule of three will be:

1L of solution has 0.80 moles of solute

Then, 0.10L of solution must have (0.1 . 0.8)/1 = 0.08 moles

This moles came from the concentrated solution, and we know that in 1L of this solution we have 1.5 moles. Therefore the rule of three will be:

1.5 moles are in 1L of solution

0.08 moles were in (0.08 . 1L / 1.5) = 0.053 L (This is the volume of concentrated solution that was used)

Dilution formula is: M conc . Vol conc = M diluted . Vol diluted

1.5 M . Vol conc = 0.80 M . 0.10L

Vol conc = 0.80 M . 0.10L / 1.5M = 0.053L

4 0

3 years ago

Read 2 more answers