Answer:
0.0847M is molarity of sodium hydrogen citrate in the solution
Explanation:
The 2.0%(w/v) solution of sodium hydrogen citrate contains 2g of the solute in 100mL of solution. To find the molarity of the solution we need to convert the mass of solute to moles using molar mass and the mL of solution to Liters because molarity is the ratio between moles of sodium hydrogen citrate and liters of solution.
<em>Moles Na2C6H6O7:</em>
<em>Molar Mass:</em>
2Na: 2*22.99g/mol: 45.98g/mol
6C: 6*12.01g/mol: 72.01g/mol
6H: 6*1.008g/mol: 6.048g/mol
7O: 7*16g/mol: 112g/mol
45.98g/mol + 72.01g/mol + 6.048g/mol + 112g/mol = 236.038g/mol
Moles of 2g:
2g * (1mol / 236.038g) = <em>8.473x10⁻³ moles</em>
<em />
<em>Liters solution:</em>
100mL * (1L / 1000mL) = <em>0.100L</em>
<em>Molarity:</em>
8.473x10⁻³ moles / 0.100L =
<h3>0.0847M is molarity of sodium hydrogen citrate in the solution</h3>
Molarity is defined as number of moles of solute in 1 L of solution.
Here, 0.1025 g of Cu is reacted with 35 mL of HNO_{3} to produced Cu^{2+} ions.
The balanced reaction will be as follows:
Cu+3HNO_{3}\rightarrow Cu(NO_{3})_{2}+NO_{2}+H_{2}O
From the above reaction, 1 mole of Cu produces 1 mole of Cu^{2+}, convert the mass of Cu into number of moles as follows:
n=\frac{m}{M}
molar mass of Cu is 63.55 g/mol thus,
n=\frac{0.1025 g}{63.55 g/mol}=0.0016 mol
Now, total molarity of solution, after addition of water is 200 mL or 0.2 L can be calculated as follows:
M=\frac{n}{V}=\frac{0.0016 mol}{0.2 L}=0.008 mol/L=0.008 M
Thus, molarity of Cu^{2+} is 0.008 M.
Answer:
Option 2, Half of the active sites are occupied by substrate
Explanation:
Michaelis-Menten expression for enzyme catalysed equation is as follows:
![V_0=\frac{V_{max\ [S]}}{k_M+[S]}](https://tex.z-dn.net/?f=V_0%3D%5Cfrac%7BV_%7Bmax%5C%20%5BS%5D%7D%7D%7Bk_M%2B%5BS%5D%7D)
Here,
is Michaelis-Menten constant and [S] is substrate concentration.
When [S]=Km
Rearrange the above equation as follows:
![V_0=\frac{V_{max}[S]}{k_M+[S]}\\V_0=\frac{V_{max}[S]}{[S]+[S]}\\V_0=\frac{V_{max}[S]}{2[S\\]}\\V_0=\frac{V_{max}}{2}](https://tex.z-dn.net/?f=V_0%3D%5Cfrac%7BV_%7Bmax%7D%5BS%5D%7D%7Bk_M%2B%5BS%5D%7D%5C%5CV_0%3D%5Cfrac%7BV_%7Bmax%7D%5BS%5D%7D%7B%5BS%5D%2B%5BS%5D%7D%5C%5CV_0%3D%5Cfrac%7BV_%7Bmax%7D%5BS%5D%7D%7B2%5BS%5C%5C%5D%7D%5C%5CV_0%3D%5Cfrac%7BV_%7Bmax%7D%7D%7B2%7D)
when [S]=Km, the rate of enzyme catalysed reaction becomes half of the maximum rate, that means half of the active sites are occupied by substrate.
Therefore, the correct option is option 2.
Answer:
1.72 L
Explanation:
Initial Temperature of the gas T =90°+273= 363 K
Initial Volume of the gas V1= 1.41 L
Final temperature of the gas T2= 170°c +273= 443 K
Final volume of the gas V2= ????
Using Charles law;
V1/T1 = V2/T2
V1T2= V2T1
V2= V1T2/T1
V2= 1.41×443/363
V2= 1.72 L