I need all solutions

An enzyme is discovered that catalyzes the chemical reaction SAD ↔ HAPPY A team of motivated researchers sets out to study the e

Pepsi [2]

Answer:

The $K_{m}$ of a substrate will be "10 μM".

Explanation:

The given values are:

$E_{t} = 20 \ nM$

$[Substract] = 40 \ \mu M$

$K_{cat}=600 \ s^{-1}$

Reaction velocity, $Vo=9.6 \ \mu M s^{-1}$

As we know,

⇒ $Vo=\frac{K_{cat}[E_{t}][S]}{K_{m}+[S]}$

On putting the estimated values, we get

⇒ $9.6=\frac{600\times 20\times 10^{-3}\times 40}{K_{m}+40}$

⇒ $K_{m}+40=\frac{600\times 20\times 10^{-3}\times 40}{9.6}$

⇒ $K_{m}+40=50$

On subtracting "40" from both sides, we get

⇒ $K_{m}+40-40=50-40$

⇒ $K_{m}=10 \ \mu M$

6 0

3 years ago

Select all the statements that correctly describe a solution. Multiple select question.

motikmotik

Answer:

B, D, C

Explanation:

the solution may have more than one solute, a solution is a homogenous mixture of 2 or more substances and not all solutions are liquid

8 0

2 years ago

Tried solving this multiple ways, but keep getting the wrong answer. So, I would appreciate the help. Thank you.

Aliun [14]

Answer:

I can give you the definition ... That might help cause I honestly don't kno the answer either ;-;

Explanation:

When used as a diacritic mark, the term dot is usually reserved for the interpunct, or to the glyphs 'combining dot above' and 'combining dot below' which may be combined with some letters of the extended Latin alphabets in use in Central European languages and Vietnamese.

Here is an example:

The dot product between a unit vector and itself is also simple to compute. In this case, the angle is zero and cosθ=1. Given that the vectors are all of length one, the dot products are i⋅i=j⋅j=k⋅k=1.

5 0

3 years ago

What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)

Genrish500 [490]

Answer:

34.28 L ( 1.5*22.4 L)

Explanation:

Calculation of the moles of aluminum as:-

Mass = 55 g

Molar mass of aluminum = 26.981539 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{55\ g}{26.981539\ g/mol}$

$Moles= 2.0384\ mol$

According to the reaction:-

$4Al+3O_2\rightarrow 2Al_2O_3$

4 moles of aluminum react with 3 moles of oxygen gas

1 mole of aluminum react with $\frac{3}{4}$ moles of oxygen gas

2.0384 moles of aluminum react with $\frac{3}{4}\times 2.0384$ moles of oxygen gas

Moles of oxygen gas = 1.5288 moles

At STP,

Pressure = 1 atm

Temperature = 273.15 K

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V = 1.5288 mol × 0.0821 L.atm/K.mol × 273.15 K

⇒V = 34.28 L ( 1.5*22.4 L)

7 0

3 years ago

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

Nina [5.8K]

<u>Answer:</u> The concentration of $NH_3$ required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

7 0

3 years ago