Question 3 of 10

HELP WHOEVER ANSWERS WILL BE MARKED BRAINLIEST

Vika [28.1K]

1) T
2) F- the heart pumps blood
3) T
4) F- atria and ventricles
5) T
6) T
7) T
8) T
9) F- your heart rate goes down but doesn't stop
10) T

1) D
2) B
3) A
4) E
5)C

7 0

2 years ago

14. An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the Km

emmainna [20.7K]

Answer:

27 min

Explanation:

The kinetics of an enzyme-catalyzed reaction can be determined by the equation of Michaelis-Menten:

$v = \frac{vmax[S]}{Km + [S]}$

Where v is the velocity in the equilibrium, vmax is the maximum velocity of the reaction (which is directed proportionally of the amount of the enzyme), Km is the equilibrium constant and [S] is the concentration of the substrate.

So, initially, the velocity of the formation of the substrate is 12μmol/9min = 1.33 μmol/min

If Km is a thousand times smaller then [S], then

v = vmax[S]/[S]

v = vmax

vmax = 1.33 μmol/min

For the new experiment, with one-third of the enzyme, the maximum velocity must be one third too, so:

vmax = 1.33/3 = 0.443 μmol/min

Km will still be much smaller then [S], so

v = vmax

v = 0.443 μmol/min

For 12 μmol formed:

0.443 = 12/t

t = 12/0.443

t = 27 min

7 0

3 years ago

For which of the following reactions is S° > 0. Choose all that apply. N2(g) + 3H2(g) 2NH3(g) 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2

Lostsunrise [7]

Answer: $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

$NH_4I(s)\rightarrow NH_3(g)+HI(g)$

$2H_2O(g)+2SO_2(g)\rightarrow 2H_2S(g)+3O_2(g)$

Explanation:

Entropy is the measure of randomness or disorder of a system.

A system has positive value of entropy if the disorder increases and a system has negative value of entropy if the disorder decreases.

1. $N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

As 4 moles of gaseous reactants are changing to 2 moles of gaseous products, the randomness is decreasing and the entropy is negative

2. $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

As 9 moles of gaseous reactants are changing to 10 moles of gaseous products, the randomness is increasing and the entropy is positive.

3. $NH_4I(s)\rightarrow NH_3(g)+HI(g)$

As 1 mole of solid reactants is changing to 2 moles of gaseous products, the randomness is increasing and the entropy is positive.

4. $2H_2O(g)+2SO_2(g)\rightarrow 2H_2S(g)+3O_2(g)$

As 4 moles of gaseous reactants is changing to 5 moles of gaseous products, the randomness is increasing and the entropy is positive

5. $2NO(g)+2H_2(g)\rightarrow N_2(g)+2H_2O(l)$

As 4 moles of gaseous reactants is changing to 1 moles of gaseous products, the randomness is decreasing and the entropy is negative.

5 0

3 years ago

How do you write a message?

Vilka [71]

Answer:

re clear. Try to convey your meaning as simply as possible. Don't over-write or use exorbitant language. ...

Are complete. Include all relevant information. Think about the situation from your readers' perspective. ...

Are correct. Always proofread before sending any message.

Explanation:

5 0

3 years ago

Read 2 more answers

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation f

dmitriy555 [2]

Answer: The mass of cryolite produced is 51.48 kg

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For aluminium oxide:

Given mass of aluminium oxide = 12.5 kg = 12500 g (Conversion factor: 1 kg = 1000 g)

Molar mass of aluminium oxide = 101.96 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium oxide}=\frac{12500g}{101.96g/mol}=122.6mol$

For NaOH:

Given mass of NaOH = 55.4 kg = 55400 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaOH}=\frac{55400g}{40g/mol}=1389mol$

For HF:

Given mass of HF = 55.4 kg = 55400 g

Molar mass of HF = 20 g/mol

Putting values in equation 1, we get:

$\text{Moles of HF}=\frac{55400g}{20g/mol}=2770mol$

For the given chemical reaction:

$Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O(g)$

By Stoichiometry of the reaction:

1 mole of aluminium oxide reacts with 6 moles of sodium hydroxide and 12 moles of HF.

So, 122.6 moles of aluminium oxide will react with $(6\times 122.6)=735.6mol$ of sodium hydroxide and $(12\times 122.6)=1471.2mol$ of HF

As, given amount of NaOH and HF is more than the required amount. So, they are considered as an excess reagent.

Thus, aluminium oxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aluminium oxide produces 2 moles of cryolite

So, 122.6 moles of aluminium oxide will produce = $\frac{2}{1}\times 122.6=245.2mol$ of cryolite

Now, calculating the mass of cryolite by using equation 1:

Molar mass of cryolite = 209.94 g/mol

Moles of cryolite = 245.2 mol

Putting values in equation 1, we get:

$245.2mol=\frac{\text{Mass of cryolite}}{209.94g/mol}\\\\\text{Mass of cryolite}=(245.2mol\times 209.94g/mol)=51477.3g$

Converting this into kilograms, we use the conversion factor:

1 kg = 1000 g

So, $51477.3 g\times (\frac{1kg}{1000g})=51.48kg$

Hence, the mass of cryolite produced is 51.48 kg

7 0

3 years ago