Answer:
Hello there! How are you doing today :] Sand would be a covalent bond.
Explanation:
Silica is one of the main compounds found in sand. It also contains many silicon and oxygen atoms. which; are joined together by covalent bonds in a regular arrangement, forming a giant covalent structure.
Answer:
The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5.
Explanation:
The first high part is Q4, then the low part is Q7, the following high part is Q6, and the energy moving from the next two high points is Q5 because of the diagram.
Answer:
The coefficient of Ca(OH)2 is 1
Explanation:
Step 1: unbalanced equation
Ca(OH)2 + HNO3 → Ca(NO3)2 + H2O
Step 2: Balancing the equation
On the right side we have 2x N (in Ca(NO3)2 ) and 1x N on the left side (in HNO3). To balance the amount of N on both sides, we have to multiply HNO3 by 2.
Ca(OH)2 + 2HNO3 → Ca(NO3)2 + H2O
On the left side we have 4x H (2xH in Ca(OH)2 and 2x H in HNO3), on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side, by 2.
Now the equationis balanced.
Ca(OH)2 + 2HNO3 = Ca(NO3)2 + 2H2O
The coefficient of Ca(OH)2 is 1
The molality of a solute is equal to the moles of solute per kg of solvent. We are given the mole fraction of I₂ in CH₂Cl₂ is <em>X</em> = 0.115. If we can an arbitrary sample of 1 mole of solution, we will have:
0.115 mol I₂
1 - 0.115 = 0.885 mol CH₂Cl₂
We need moles of solute, which we have, and must convert our moles of solvent to kg:
0.885 mol x 84.93 g/mol = 75.2 g CH₂Cl₂ x 1 kg/1000g = 0.0752 kg CH₂Cl₂
We can now calculate the molality:
m = 0.115 mol I₂/0.0752 kg CH₂Cl₂
m = 1.53 mol I₂/kg CH₂Cl₂
The molality of the iodine solution is 1.53.
Use PV = nRT
(2 atm)(.3 liters) = n(8.314 mol*K)(303°K)
.6 = n(2519.142)
Divide by 2519.142
n = .00023818 mols of HCl * 36.46g of HCl/ 1 mol of HCl
Grams of HCl = 0.00868
