molar mass of H2SO4= 2(1)+ 32 + 4(16)
=98g/mol
no. of moles = no. of molecules/6.02*10^23
no. of moles = 2.05*10^16/6.02*10^23
no. of moles = 3.406*10^(-8)mol
mass of a substance = molar mass*no. of moles
= 98*3.406*10^(-8)
=3.34*10^-6g Ans
To find the mass of glucose in the solution, we first need to calculate the volume of glucose in the solution. Using the formula m = V x mv, we find that the mass of glucose (m) in 205 mL of a 5.50 mv glucose solution is approximately 11.28 grams.
There are approximately 0.451 pounds in 205 grams.
To calculate the grams of NaOH needed, you first need to find the moles of NaOH required: 205 M concentration means 205 moles/L. So, for 250 mL (0.25 L), you multiply 0.25 L by 205 moles/L to get 51.25 moles. Finally, using the molar mass of NaOH (40 g/mol), you can convert moles to grams by multiplying 51.25 moles by 40 g/mol to find you need 2050 grams of NaOH.
It will have about 150-160 hp unless it is supercharged. It could possibly be 205 hp
The thermal conductivity of aluminum is about 205 watts per meter-kelvin (W/mK).
the overall atomic mass is 204.37, but for thallium 203 it is 203, and for thallium 205, it is 205.
To find the number of moles present in 205 g of helium, we need to use the molar mass of helium, which is approximately 4 g/mol. Divide the given mass by the molar mass to get the number of moles. In this case, 205 g / 4 g/mol = 51.25 moles of helium.
(203 x 0.30) = 60.9 (205 x 0.70) = 143.5 60.9 + 143.5 = 204.4 The answer is 204.4 amu !
205
30 percent of 205 = 61.530% of 205= 30% * 205= 0.30 * 205= 61.5
The factors of 205 are: 1 5 41 205
10% of 205:= 10% x 205= 0.10 x 205= 20.5
The factors of 205 are 1, 5, 41, and 205.
The factors of 205 are: 1 5 41 205
The factors of 205 are: 1, 5, 41, 205
410, 615, 820, 1025, 1230; 205*2=410 205*3=615 205*4=820 205*5=1025 205*6=1230
=205-200/205*100