9.32*1024 (molec's CH3OH) / 6.022*1023 (molec's.mol−1 CH3OH) * 32.04 (g mol−1 CH3OH) = 495.8 g = 496 g CH3OH
To calculate the mass of methanol (CH3OH), we first need to find the molar mass of CH3OH. The molar mass of CH3OH is approximately 32 grams per mole. By multiplying the molar mass by the given amount of 9.85x10^24 molecules, we can find the mass in grams, which is about 3.15x10^26 grams.
To find the mass of 9.03 x 10^24 molecules of methanol (CH3OH), we first calculate the molar mass of CH3OH: (1 x 12.01 g/mol) + (4 x 1.01 g/mol) + (1 x 16.00 g/mol) = 32.04 g/mol Then we can convert the number of molecules to moles and finally to grams: 9.03 x 10^24 molecules * (1 mol / 6.022 x 10^23 molecules) * 32.04 g/mol ≈ 482 g
I am assuming that the question is: what is the mass, in grams, of 9.93x1024 molecules of methanol (CH3OH). For any question concerning mass and a chemical, we must find the molecular mass by adding the atomic masses from the periodic table: C=12.0g 4H= 4 x 1.01 = 4.04g O=16.0g Total: 32.0g/mole Now, how many moles is 9.93x1024 molecules, if one mole is 6.02x1023? (9.93x1024 ) / (6.02x1023) = 15.5 moles. So, 15.5 moles x 32.0g/mole = 496g.
To prepare a 1.5 M solution of CH3OH in 150 mL, you need to calculate the moles required first. 1.5 moles/L * 0.15 L = 0.225 moles of CH3OH. Since the molar mass of CH3OH is 32 g/mol, you would need 7.2 grams of CH3OH to prepare the solution.
To find the mass of 3.62 x 10^24 molecules of CH3OH, you need to first calculate the molar mass of CH3OH, which is 32.04 g/mol. Then, you can use Avogadro's number (6.022 x 10^23 molecules/mol) to convert the number of molecules to moles. Finally, multiply the number of moles by the molar mass to find the mass.
To calculate the mass of methanol (CH3OH), we first need to find the molar mass of CH3OH. The molar mass of CH3OH is approximately 32 grams per mole. By multiplying the molar mass by the given amount of 9.85x10^24 molecules, we can find the mass in grams, which is about 3.15x10^26 grams.
9.18x10^24 molecules CH3OH x 1 mole/6.02x10^23 molecules x 32 g/mole = 488 g (to 3 sig figs)
To find the mass of 9.03 x 10^24 molecules of methanol (CH3OH), we first calculate the molar mass of CH3OH: (1 x 12.01 g/mol) + (4 x 1.01 g/mol) + (1 x 16.00 g/mol) = 32.04 g/mol Then we can convert the number of molecules to moles and finally to grams: 9.03 x 10^24 molecules * (1 mol / 6.022 x 10^23 molecules) * 32.04 g/mol ≈ 482 g
To find the mass of 9.70 × 10²⁴ molecules of methanol (CH₃OH), first calculate the number of moles using Avogadro's number (6.022 × 10²³ molecules/mol). This gives approximately 16.13 moles of methanol. The molar mass of methanol is about 32.04 g/mol, so the total mass is 16.13 moles × 32.04 g/mol ≈ 516.6 grams.
To find the mass of 9.52 × 10²⁴ molecules of methanol (CH₃OH), first calculate the number of moles using Avogadro's number (6.022 × 10²³ molecules/mol). This gives approximately 15.8 moles of methanol. The molar mass of methanol is about 32.04 g/mol, so the mass is calculated by multiplying the number of moles by the molar mass, resulting in approximately 506 grams.
To find the mass of 9.39 × 10²⁴ molecules of methanol (CH₃OH), first determine the number of moles using Avogadro's number (6.022 × 10²³ molecules/mol). Calculate the moles: 9.39 × 10²⁴ molecules ÷ 6.022 × 10²³ molecules/mol ≈ 15.59 moles. The molar mass of methanol is approximately 32.04 g/mol, so the mass is 15.59 moles × 32.04 g/mol ≈ 499.5 grams.
The mass is 483,62 g.
8.20g of carbon monoxide, how many grams of methanol will be produced?
I am assuming that the question is: what is the mass, in grams, of 9.93x1024 molecules of methanol (CH3OH). For any question concerning mass and a chemical, we must find the molecular mass by adding the atomic masses from the periodic table: C=12.0g 4H= 4 x 1.01 = 4.04g O=16.0g Total: 32.0g/mole Now, how many moles is 9.93x1024 molecules, if one mole is 6.02x1023? (9.93x1024 ) / (6.02x1023) = 15.5 moles. So, 15.5 moles x 32.0g/mole = 496g.
The answer is 50,196 g.
The simplest form of the molecular formula for methanol {note corrected spelling} is CH4O, and its gram molecular mass is 32.04. By definition therefore, a mass of Avogadro's Number of molecules contains 32.04 grams. Avogadro's Number is about 6.022 X 1023. Therefore, 9.47 X 1024 molecules of methanol contains [(9.47 X 1024)/(6.022 X 1023)]32.04 or 504 grams, to the justified number of significant digits.
To find the mass of 9.36 x 10²⁴ molecules of methanol (CH₃OH), we first need to determine the number of moles. Using Avogadro's number (6.022 x 10²³ molecules/mol), we calculate: [ \text{Moles of CH₃OH} = \frac{9.36 \times 10^{24} \text{ molecules}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 15.55 \text{ moles} ] The molar mass of methanol is approximately 32.04 g/mol. Therefore, the mass is: [ \text{Mass} = 15.55 \text{ moles} \times 32.04 \text{ g/mol} \approx 498.6 \text{ grams} ] Thus, the mass of 9.36 x 10²⁴ molecules of methanol is approximately 498.6 grams.