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134.1 grams
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How many milligrams are equivalent to 0.045 grams?
0.045 grams is 45 milligrams.
How many atoms are in 45 grams of pb?
First of all we convert the mass of leadto moles. Using the equation moles = mass(g) / Ar (Realtive atomic Mass)_. Mass( g) = 45 g Ar =~ 207 (Periodic Table). Hence moles(Pb) = 45 x 207 Moles (Pb) = 9315 moles. Next using the Avogadro number. 1 moles(of A SUBSTANCE) contains 6.022 x 10^(23) atoms. Hence number of atoms in 9315 moles = 9315 x 6.022 x 10^(23) = 5.61 x 10^(27) atoms. As a 'silly' number. 5,610,000,000,000,000,000,000,000,000 atoms.
How many molecules are in 4.21 grams of water?
45 grams H20 x (1 mole H20/18 grams H2O) x (6.02E23 molecules H20/1 mole H2O) the grams H2O and moles H2O cancel out. When you punch it into your calculator, the answer comes out to: =1.505E24 molecules H2O
How many grams are in 1.6 oz?
16oz = 453.6g
What is the mass in grams of 45 ml of water?
45mL of plain water has a mass of about 45 grams.
How many moles are in 45 grams of potassium oxide?
45/94.2 is 0.4777 moles
How many grams of zinc chloride does it take to react to 45 grams of aluminum?
To determine how many grams of zinc chloride (ZnCl₂) are needed to react with 45 grams of aluminum (Al), we first need to look at the balanced chemical equation for the reaction, which is: [ 2Al + 3ZnCl₂ \rightarrow 2AlCl₃ + 3Zn ] From the equation, 2 moles of aluminum react with 3 moles of zinc chloride. The molar mass of aluminum is approximately 27 g/mol, so 45 grams of aluminum corresponds to about 1.67 moles of Al. Thus, we need 2.5 moles of ZnCl₂ (since 1.67 moles Al will require 2.5 moles ZnCl₂). The molar mass of ZnCl₂ is about 136.3 g/mol, so 2.5 moles of ZnCl₂ would weigh approximately 340.75 grams.
How many grams are in 2.50 moles of H2O?
To calculate the number of grams in 2.50 moles of H2O, you first need to know the molar mass of water (H2O), which is approximately 18.015 g/mol. Then, you can use the formula: grams = moles x molar mass. So, 2.50 moles of H2O is equivalent to 45.04 grams (2.50 moles x 18.015 g/mol).
How many grams of carbon dioxide would be produced if 45 grams of glucose completely react with oxygen?
The chemical equation for the combustion of glucose is C6H12O6 + 6O2 -> 6CO2 + 6H2O. From this, we can see that 1 mole of glucose produces 6 moles of CO2. Therefore, 45 grams of glucose (which is approximately 0.25 moles) would produce 6 times that amount of CO2, which is about 90 grams.
How many grams of carbon dioxide would be produced if 45 g of C6H12O6 competely reacted with oxygen?
C6H12O6 + 6O2 --> 6CO2 + 6H2O 45 grams C6H12O6 (1 mole C6H12O6/180.156 grams)(6 moles CO2/1 mole C6H12O6)(44.01 grams/1 mole CO2) = 66 grams carbon dioxide produced ==========================
45 grams is equal to how many kilograms?
There's a thousand grams in a kilogram, 45 grams is 0.045 kilograms.
How many milligrams are equivalent to 0.045 grams?
0.045 grams is 45 milligrams.
How many grams are there in 45 percent of a kilogram?
450 grams
How many atoms are in 45 grams of pb?
First of all we convert the mass of leadto moles. Using the equation moles = mass(g) / Ar (Realtive atomic Mass)_. Mass( g) = 45 g Ar =~ 207 (Periodic Table). Hence moles(Pb) = 45 x 207 Moles (Pb) = 9315 moles. Next using the Avogadro number. 1 moles(of A SUBSTANCE) contains 6.022 x 10^(23) atoms. Hence number of atoms in 9315 moles = 9315 x 6.022 x 10^(23) = 5.61 x 10^(27) atoms. As a 'silly' number. 5,610,000,000,000,000,000,000,000,000 atoms.
How many grams are there in 45 kg?
1kg = 1000g. Therefore, 45 x 1000 = 45000 grams.
How many grams per liter in 6.25 kg to 45 liters?
6.25 kg/45 litres = 6250 grams/45 litres = 6250/45 grams per litre =138.88... gpl
How many molecules are there in 45 grams of aluminum triflouride?
To find the number of molecules in 45 grams of aluminum trifluoride (AlF₃), we first need to calculate the molar mass of AlF₃, which is approximately 26.98 g/mol (Al) + 3 × 19.00 g/mol (F) = 83.98 g/mol. Next, we can determine the number of moles in 45 grams by dividing the mass by the molar mass: 45 g / 83.98 g/mol ≈ 0.536 moles. Finally, using Avogadro's number (approximately (6.022 \times 10^{23}) molecules/mol), we find the number of molecules: 0.536 moles × (6.022 \times 10^{23}) molecules/mol ≈ (3.22 \times 10^{23}) molecules of AlF₃.
