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What is the mass (in grams) of 9.52 1024 molecules of methanol (CH3OH)?
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.
What is the mass in grams of 9.70 1024 molecules of methanol CH3OH?
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.
What is the mass in grams of 9.39 and times1024 molecules of methanol (CH3OH)?
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.
What is the mass in grams of 9.71 x 1024 molecules of methanol?
To find the mass of 9.71 x 10²⁴ molecules of methanol (CH₃OH), we first need to determine the molar mass of methanol, which is approximately 32.04 g/mol. Next, we use Avogadro's number (6.022 x 10²³ molecules/mol) to convert molecules to moles: 9.71 x 10²⁴ molecules is about 16.14 moles. Finally, multiplying the number of moles by the molar mass gives: 16.14 moles x 32.04 g/mol ≈ 517.6 grams.
What is the mass in grams of 9.36 1024 molecules of methanol CH3OH?
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.
What is the mass(in grams) of 9.401024 molecules of methanol?
The answer is 50,196 g.
What is the mass (in grams) of 9.85x10 to the 24 of methanol (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.
What is the mass (in grams) of 9.52 1024 molecules of methanol (CH3OH)?
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.
What is the mass in grams of 9.39 and times1024 molecules of methanol (CH3OH)?
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.
What is the mass in grams of 9.47 x 1024 molecules of methonal?
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.
What is the mass in grams of 9.71 x 1024 molecules of methanol?
To find the mass of 9.71 x 10²⁴ molecules of methanol (CH₃OH), we first need to determine the molar mass of methanol, which is approximately 32.04 g/mol. Next, we use Avogadro's number (6.022 x 10²³ molecules/mol) to convert molecules to moles: 9.71 x 10²⁴ molecules is about 16.14 moles. Finally, multiplying the number of moles by the molar mass gives: 16.14 moles x 32.04 g/mol ≈ 517.6 grams.
What is the mass in grams of 9.32x1024 molecules of methanol CH3OH?
To calculate the mass of 9.32x10^24 molecules of methanol (CH3OH), you can first find the molar mass of CH3OH, which is approximately 32 g/mol. Then, you can convert the number of molecules to moles and finally to grams. The calculation would be (9.32x10^24 molecules) / (6.022x10^23 molecules/mol) * (32 g/mol) = 497 grams.
What is the mass in grams of 9.93x1024 molecules of methanol CH3OH?
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.
What is the mass in grams of 9.15 1024 molecules of methanol CH3OH?
To find the mass of 9.15 × 10²⁴ molecules of methanol (CH₃OH), we first need to determine the molar mass of methanol, which is approximately 32.04 g/mol. Using Avogadro's number (6.022 × 10²³ molecules/mol), we can convert the number of molecules to moles: ( n = \frac{9.15 \times 10^{24}}{6.022 \times 10^{23}} \approx 15.19 ) moles. Finally, we calculate the mass: ( \text{mass} = n \times \text{molar mass} = 15.19 , \text{mol} \times 32.04 , \text{g/mol} \approx 486.4 , \text{g} ).
What is the mass (in grams) of 9.18 1024 molecules of methanol (CH3OH)?
9.18x10^24 molecules CH3OH x 1 mole/6.02x10^23 molecules x 32 g/mole = 488 g (to 3 sig figs)
What mass of methanol (CH3OH) is produced when 140.1 g of carbon monoxide reacts with 12.12 g of hydrogen?
8.20g of carbon monoxide, how many grams of methanol will be produced?
What is the mass in grams of 1.20x1025 molecules of ammonia NH3?
To find the mass in grams of 1.20x10^25 molecules of ammonia (NH3), you first calculate the molar mass of NH3 (17.031 g/mol). Then, divide the given number of molecules by Avogadro's number (6.022x10^23 molecules/mol) to find the number of moles, and finally, multiply the number of moles by the molar mass to get the mass in grams, which will be approximately 4.08x10^2 grams.
