Answer:
The minimum molecular weight of the enzyme is 29.82 g/mol
Explanation:
<u>Step 1:</u> Given data
The volume of the solution = 10 ml = 10*10^-3L
Molarity of the solution = 1.3 mg/ml
moles of AgNO3 added = 0.436 µmol = 0.436 * 10^-3 mmol
<u>Step 2:</u> Calculate the mass
Density = mass/ volume
1.3mg/mL = mass/ 10.0 mL
mass = 1.3mg/mL *10.0 mL = 13mg
<u>Step 3:</u> Calculate minimum molecular weight
Molecular weight = mass of the enzyme / number of moles
Molecular weight of the enzyme = 13mg/ 0.436 * 10^-3 mmol
Molecular weight = 29.82 g/mole
The minimum molecular weight of the enzyme is 29.82 g/mol
The much of the sample that would remain unchanged after 140 seconds is 2.813 g
Explanation
Half life is time taken for the quantity to reduce to half its original value.
if the half life for Scandium is 35 sec, then the number of half life in 140 seconds
=140 sec/ 35 s = 4 half life
Therefore 45 g after first half life = 45 x1/2 =22.5 g
22.5 g after second half life = 22.5 x 1/2 =11.25 g
11.25 g after third half life = 11.25 x 1/2 = 5.625 g
5.625 after fourth half life = 5.625 x 1/2 = 2.813
therefore 2.813 g of Scandium 47 remains unchanged.
Answer:
0.302 moles
Explanation:
Data given
Mass of Pb(NO₃)₂ = 100 g
Moles of Pb(NO₃)₂ = ?
Solution:
To find mole we have to know about molar mass of Pb(NO₃)₂
So,
Molar mass of Pb(NO₃)₂ = 207 + 2[14 + 3(16)]
= 207 + 2[14 + 48]
= 207 + 124
Molar mass of Pb(NO₃)₂ = 331 g/mol
Formula used :
no. of moles = mass in grams / molar mass
Put values in above formula
no. of moles = 100 g / 331 g/mol
no. of moles = 0.302 moles
no. of moles of Pb(NO₃)₂ = 0.302 moles
Explanation:
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have not yet learnt chemistry so sorry
Answer:
P = 0.6815 atm
Explanation:
Pressure = 754 torr
The conversion of P(torr) to P(atm) is shown below:
So,
Pressure = 754 / 760 atm = 0.9921 atm
Temperature = 294 K
Volume = 3.1 L
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:
0.9921 atm × 3.1 L = n × 0.0821 L.atm/K.mol × 294 K
⇒n of helium gas= 0.1274 moles
Surface are = 1257 cm²
For a sphere, Surface area = 4 × π × r² = 1257 cm²
r² = 1257 / 4 × π ≅ 100 cm²
r = 10 cm
The volume of the sphere is :
Where, V is the volume
r is the radius
V = 4190.4762 cm³
1 cm³ = 0.001 L
So, V (max) = 4.19 L
T = 273 K
n = 0.1274 moles
Using ideal gas equation as:
PV=nRT
Applying the equation as:
P × 4.19 L = 0.1274 × 0.0821 L.atm/K.mol × 273 K
<u>P = 0.6815 atm</u>
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