0
What else can I help you with?
How many moles is 2.80 times 10 to the twenty third power atoms of silicon?
The answer is 0,465 moles.
How many uranium atoms do you have with 0.70 moles of uranium atoms?
To find the number of uranium atoms in 0.70 moles, you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) atoms per mole. Multiplying the number of moles by Avogadro's number gives: (0.70 , \text{moles} \times 6.022 \times 10^{23} , \text{atoms/mole} \approx 4.21 \times 10^{23} , \text{atoms}). Thus, there are approximately (4.21 \times 10^{23}) uranium atoms in 0.70 moles.
How many moles of nickel atoms are there in 8.00 x 109 Ni atoms?
To find the number of moles of nickel atoms in (8.00 \times 10^9) Ni atoms, you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) atoms/mole. The calculation is as follows: [ \text{Moles of Ni} = \frac{8.00 \times 10^9 \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mole}} \approx 1.33 \times 10^{-14} \text{ moles} ] Thus, there are approximately (1.33 \times 10^{-14}) moles of nickel atoms in (8.00 \times 10^9) Ni atoms.
How many moles of nitrogen are contained in 1.61 x 10 atoms of nitrogen?
To find the number of moles of nitrogen in (1.61 \times 10^{24}) atoms, you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) atoms per mole. Calculating the moles: [ \text{Moles of nitrogen} = \frac{1.61 \times 10^{24} \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mole}} \approx 2.68 \text{ moles} ] Thus, there are approximately 2.68 moles of nitrogen in (1.61 \times 10^{24}) atoms.
How many moles are in 1.63x10 to the 24th atoms?
To find the number of moles in (1.63 \times 10^{24}) atoms, you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) atoms per mole. Divide the number of atoms by Avogadro's number: [ \text{moles} = \frac{1.63 \times 10^{24}}{6.022 \times 10^{23}} \approx 2.71 \text{ moles}. ] Thus, there are approximately 2.71 moles in (1.63 \times 10^{24}) atoms.
How many moles of Krypton are equivalent to 1.3 times 10 to the 16th power atoms of Kr?
Since a molecule of krypton is a single atom, the answer to this question is the quotient of 1.3 X 1016 divided by Avogadro's Number, or about 2.2 X 10-8 mole.
How many moles is 2.80 times 10 to the twenty third power atoms of silicon?
The answer is 0,465 moles.
How many uranium atoms do you have with 0.70 moles of uranium atoms?
To find the number of uranium atoms in 0.70 moles, you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) atoms per mole. Multiplying the number of moles by Avogadro's number gives: (0.70 , \text{moles} \times 6.022 \times 10^{23} , \text{atoms/mole} \approx 4.21 \times 10^{23} , \text{atoms}). Thus, there are approximately (4.21 \times 10^{23}) uranium atoms in 0.70 moles.
How many moles of nickel atoms are there in 8.00 x 109 Ni atoms?
To find the number of moles of nickel atoms in (8.00 \times 10^9) Ni atoms, you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) atoms/mole. The calculation is as follows: [ \text{Moles of Ni} = \frac{8.00 \times 10^9 \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mole}} \approx 1.33 \times 10^{-14} \text{ moles} ] Thus, there are approximately (1.33 \times 10^{-14}) moles of nickel atoms in (8.00 \times 10^9) Ni atoms.
How many moles of nitrogen are contained in 1.61 x 10 atoms of nitrogen?
To find the number of moles of nitrogen in (1.61 \times 10^{24}) atoms, you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) atoms per mole. Calculating the moles: [ \text{Moles of nitrogen} = \frac{1.61 \times 10^{24} \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mole}} \approx 2.68 \text{ moles} ] Thus, there are approximately 2.68 moles of nitrogen in (1.61 \times 10^{24}) atoms.
How many moles are in 1.63x10 to the 24th atoms?
To find the number of moles in (1.63 \times 10^{24}) atoms, you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) atoms per mole. Divide the number of atoms by Avogadro's number: [ \text{moles} = \frac{1.63 \times 10^{24}}{6.022 \times 10^{23}} \approx 2.71 \text{ moles}. ] Thus, there are approximately 2.71 moles in (1.63 \times 10^{24}) atoms.
How many atoms of magnesium in 2.5 moles of magnesium?
To find the number of atoms in 2.5 moles of magnesium, you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) atoms per mole. Multiplying 2.5 moles by Avogadro's number gives you: [ 2.5 , \text{moles} \times 6.022 \times 10^{23} , \text{atoms/mole} \approx 1.51 \times 10^{24} , \text{atoms}. ] Therefore, there are about (1.51 \times 10^{24}) atoms of magnesium in 2.5 moles.
How many atoms are in 1.2 moles U?
To find the number of atoms in 1.2 moles of uranium (U), you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) atoms per mole. Multiply the number of moles by Avogadro's number: [1.2 , \text{moles} \times 6.022 \times 10^{23} , \text{atoms/mole} \approx 7.23 \times 10^{23} , \text{atoms}.] Thus, there are approximately (7.23 \times 10^{23}) atoms in 1.2 moles of uranium.
Number of moles of hydrogen atoms in 10.7 mol water?
H2O is water. One mole of water contains 2 moles of hydrogen atoms. Therefore, 10.7 moles of water contain 21.4 moles of hydrogen atom.
How many atoms are in 4 moles of lithium?
To find the number of atoms in 4 moles of lithium, you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) atoms per mole. Therefore, in 4 moles of lithium, the number of atoms is (4 \times 6.022 \times 10^{23} = 2.409 \times 10^{24}) atoms.
How many atoms are in 6.2 moles of Al?
To find the number of atoms in 6.2 moles of aluminum (Al), you can use Avogadro's number, which is approximately (6.022 \times 10^{23}) atoms per mole. Therefore, the calculation is (6.2 , \text{moles} \times 6.022 \times 10^{23} , \text{atoms/mole} \approx 3.74 \times 10^{24} , \text{atoms}). Thus, there are approximately (3.74 \times 10^{24}) atoms in 6.2 moles of Al.
How many nickel atoms are present in 5.6 moles of nickel metal?
Avogadro's number times 5.6 (or about 33.6 times ten to the 23rd power.
