CUSO was created in 1961.
To determine the limiting reagent between CuSO₄ and NaOH, we first need to look at the balanced chemical equation for the reaction, which is: [ CuSO₄ + 2 NaOH \rightarrow Cu(OH)₂ + Na₂SO₄ ] According to the stoichiometry, 1 mole of CuSO₄ reacts with 2 moles of NaOH. First, calculate the moles of CuSO₄ and NaOH. Assuming the molar mass of CuSO₄ is approximately 159.61 g/mol, 638.44 g of CuSO₄ corresponds to about 4.00 moles. For 240.0 g of NaOH (molar mass ≈ 40.00 g/mol), this is about 6.00 moles. Since 4.00 moles of CuSO₄ would require 8.00 moles of NaOH but only 6.00 moles are available, NaOH is the limiting reagent.
To determine how many grams of CuSO₄ are required, we first need to calculate the moles of KI. With 300.0 mL of 0.75 mol/L KI, the moles of KI are 0.300 L × 0.75 mol/L = 0.225 moles. Assuming a 1:1 reaction ratio (which is typical for reactions between KI and CuSO₄), 0.225 moles of CuSO₄ are needed. The molar mass of CuSO₄ is approximately 159.61 g/mol, so the mass required is 0.225 moles × 159.61 g/mol = 35.9 grams of CuSO₄.
CUSO is not a recognized chemical element. It may refer to a chemical compound like copper sulfate (CuSO4), which is a blue crystalline solid commonly used in agriculture and industry.
Copper sulfate (CuSO₄) contains copper (Cu) in the +2 oxidation state. In this state, copper has an electron configuration of [Ar] 3d⁹, resulting in one unpaired electron in the 3d subshell. Therefore, the expected number of unpaired electrons in CuSO₄ is one.
One molecule of copper sulfate (CuSO₄) contains a total of five atoms. This includes one copper atom (Cu), one sulfur atom (S), and four oxygen atoms (O), resulting in a chemical formula of CuSO₄.
CUSO-VSO was created in 1961.
The answer is stoeker whale!
CuSO 4
To determine the limiting reagent between CuSO₄ and NaOH, we first need to look at the balanced chemical equation for the reaction, which is: [ CuSO₄ + 2 NaOH \rightarrow Cu(OH)₂ + Na₂SO₄ ] According to the stoichiometry, 1 mole of CuSO₄ reacts with 2 moles of NaOH. First, calculate the moles of CuSO₄ and NaOH. Assuming the molar mass of CuSO₄ is approximately 159.61 g/mol, 638.44 g of CuSO₄ corresponds to about 4.00 moles. For 240.0 g of NaOH (molar mass ≈ 40.00 g/mol), this is about 6.00 moles. Since 4.00 moles of CuSO₄ would require 8.00 moles of NaOH but only 6.00 moles are available, NaOH is the limiting reagent.
To determine how many grams of CuSO₄ are required, we first need to calculate the moles of KI. With 300.0 mL of 0.75 mol/L KI, the moles of KI are 0.300 L × 0.75 mol/L = 0.225 moles. Assuming a 1:1 reaction ratio (which is typical for reactions between KI and CuSO₄), 0.225 moles of CuSO₄ are needed. The molar mass of CuSO₄ is approximately 159.61 g/mol, so the mass required is 0.225 moles × 159.61 g/mol = 35.9 grams of CuSO₄.
CUSO is not a recognized chemical element. It may refer to a chemical compound like copper sulfate (CuSO4), which is a blue crystalline solid commonly used in agriculture and industry.
how should anyone know? maybe albert ienstien
CuSO₄+2NaOH=Na₂SO₄+Cu(OH)₂↓
Copper sulfate (CuSO₄) contains copper (Cu) in the +2 oxidation state. In this state, copper has an electron configuration of [Ar] 3d⁹, resulting in one unpaired electron in the 3d subshell. Therefore, the expected number of unpaired electrons in CuSO₄ is one.
One molecule of copper sulfate (CuSO₄) contains a total of five atoms. This includes one copper atom (Cu), one sulfur atom (S), and four oxygen atoms (O), resulting in a chemical formula of CuSO₄.
20)=5.18=90(g)
To find the mass of water that can be removed from 3.30 moles of CuSO₄·5H₂O, we first note that each formula unit of CuSO₄·5H₂O contains 5 moles of water. Therefore, 3.30 moles of CuSO₄·5H₂O contains (3.30 \times 5 = 16.5) moles of water. Since the molar mass of water (H₂O) is approximately 18 grams per mole, the mass of water that can be removed is (16.5 \times 18 \approx 297) grams.