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For this you need the atomic (molecular) mass of C6H5OH. Take the number of grams and divide it by the Atomic Mass. Multiply by one mole for units to cancel. C6H5OH=94.1 grams
25.5 grams C6H5OH / (94.1 grams) = .271 moles C6H5OH
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If you have 125g of zinc how many moles of zinc do you have?
To find the number of moles of zinc, you can use the formula: moles = mass (g) / molar mass (g/mol). The molar mass of zinc is approximately 65.38 g/mol. Thus, for 125 g of zinc, the calculation would be 125 g / 65.38 g/mol, which equals approximately 1.91 moles of zinc.
If you have 125 grams of zinc how many moles of zinc do you have?
To calculate the number of moles of zinc, use the formula: moles = mass (grams) / molar mass (grams/mol). The molar mass of zinc (Zn) is approximately 65.38 g/mol. Therefore, 125 grams of zinc is equal to 125 g / 65.38 g/mol, which is about 1.91 moles of zinc.
How many moles of mercury (II) oxide are needed to produce 125 g of oxygen?
To determine how many moles of mercury (II) oxide (HgO) are needed to produce 125 g of oxygen (O₂), we first need to consider the decomposition reaction: 2 HgO(s) → 2 Hg(l) + O₂(g). From this equation, we see that 2 moles of HgO produce 1 mole of O₂. The molar mass of O₂ is approximately 32 g/mol, so 125 g of O₂ corresponds to about 3.91 moles (125 g ÷ 32 g/mol). Therefore, since 2 moles of HgO produce 1 mole of O₂, we need 7.82 moles of HgO (3.91 moles O₂ × 2 moles HgO/mole O₂).
How many moles of mercury (II) oxide are needed to produce 125 grams of oxygen?
To find the moles of mercury (II) oxide (HgO) needed to produce 125 grams of oxygen (O2), we first calculate the moles of O2. The molar mass of O2 is approximately 32 g/mol, so 125 g of O2 corresponds to about 3.91 moles (125 g ÷ 32 g/mol). The decomposition of 2 moles of HgO produces 1 mole of O2, meaning we need 7.82 moles of HgO (3.91 moles O2 × 2) to produce that amount of oxygen. Thus, 7.82 moles of mercury (II) oxide are required.
How many moles of mercury are produced if 125 g of oxygen is also produced?
To determine how many moles of mercury are produced when 125 g of oxygen is generated, we first need to know the balanced chemical equation for the reaction involving mercury and oxygen. Assuming the reaction is the formation of mercury(II) oxide (HgO) from mercury (Hg) and oxygen (O₂), the equation is: 2 Hg + O₂ → 2 HgO. Given that the molar mass of oxygen (O₂) is approximately 32 g/mol, 125 g of oxygen corresponds to about 3.91 moles of O₂. According to the stoichiometry of the balanced equation, 2 moles of Hg are produced for every 1 mole of O₂. Therefore, 3.91 moles of O₂ would produce approximately 7.82 moles of Hg.
How many moles of nickel (Ni) are there in 125 g of Ni?
125 g nickel is equivalent to 2,13 moles.
How many moles of nickel (Ni) atoms are in 125 g Ni?
To find the number of moles of nickel atoms in 125 g of nickel, divide the given mass by the molar mass of nickel. The molar mass of nickel is approximately 58.69 g/mol. Therefore, 125 g Ni / 58.69 g/mol = ~2.13 moles of Ni atoms.
If you have 125g of zinc how many moles of zinc do you have?
To find the number of moles of zinc, you can use the formula: moles = mass (g) / molar mass (g/mol). The molar mass of zinc is approximately 65.38 g/mol. Thus, for 125 g of zinc, the calculation would be 125 g / 65.38 g/mol, which equals approximately 1.91 moles of zinc.
If you have 125 grams of zinc how many moles of zinc do you have?
To calculate the number of moles of zinc, use the formula: moles = mass (grams) / molar mass (grams/mol). The molar mass of zinc (Zn) is approximately 65.38 g/mol. Therefore, 125 grams of zinc is equal to 125 g / 65.38 g/mol, which is about 1.91 moles of zinc.
How many moles of mercury (II) oxide are needed to produce 125 g of oxygen?
To determine how many moles of mercury (II) oxide (HgO) are needed to produce 125 g of oxygen (O₂), we first need to consider the decomposition reaction: 2 HgO(s) → 2 Hg(l) + O₂(g). From this equation, we see that 2 moles of HgO produce 1 mole of O₂. The molar mass of O₂ is approximately 32 g/mol, so 125 g of O₂ corresponds to about 3.91 moles (125 g ÷ 32 g/mol). Therefore, since 2 moles of HgO produce 1 mole of O₂, we need 7.82 moles of HgO (3.91 moles O₂ × 2 moles HgO/mole O₂).
How many moles of mercury (II) oxide are needed to produce 125 grams of oxygen?
To find the moles of mercury (II) oxide (HgO) needed to produce 125 grams of oxygen (O2), we first calculate the moles of O2. The molar mass of O2 is approximately 32 g/mol, so 125 g of O2 corresponds to about 3.91 moles (125 g ÷ 32 g/mol). The decomposition of 2 moles of HgO produces 1 mole of O2, meaning we need 7.82 moles of HgO (3.91 moles O2 × 2) to produce that amount of oxygen. Thus, 7.82 moles of mercury (II) oxide are required.
How many moles are in 125g of Zn?
The equivalent of 125 g zinc is 1, 91 moles.
How many moles of mercury are produced if 125 g of oxygen is also produced?
To determine how many moles of mercury are produced when 125 g of oxygen is generated, we first need to know the balanced chemical equation for the reaction involving mercury and oxygen. Assuming the reaction is the formation of mercury(II) oxide (HgO) from mercury (Hg) and oxygen (O₂), the equation is: 2 Hg + O₂ → 2 HgO. Given that the molar mass of oxygen (O₂) is approximately 32 g/mol, 125 g of oxygen corresponds to about 3.91 moles of O₂. According to the stoichiometry of the balanced equation, 2 moles of Hg are produced for every 1 mole of O₂. Therefore, 3.91 moles of O₂ would produce approximately 7.82 moles of Hg.
If you have g of beryllium how many moles is this equivalent to?
The formula is: number of moles = g Be/9,012.
What is the mass of nitrogen in 125 g of ammonia?
In ammonia (NH3), the molar mass is 17 g/mol. To find the mass of nitrogen in 125 g of ammonia, first, calculate the number of moles of ammonia in 125 g. Then, multiply the moles of ammonia by the molar ratio of nitrogen in ammonia (1 mol of N for every 1 mol of NH3), and finally, multiply by the molar mass of nitrogen (14 g/mol) to find the mass of nitrogen. This will give the mass of nitrogen in 125 g of ammonia.
How many moles are in 14.84 g Mg?
14,84 g magnesium are equivalent to 0,61 moles.
How many moles are in 97.5 g of oxygen?
97,5 g of oxygen is equal to 5,416 moles.
