Moles He = 7.83 x 10^24 / 6.02 x 10^23 =13.0
<span>mass He = 13.0 mol x 4.00 g/mol = 52.0 g</span>
<span>1. 1 molecule of C6H12O6(dextrose sugar), 2 molecles of c2h6o (ethyl alcohol), 2 molecules of Co2
2. 48 hydrogen atoms </span>
Answer: It's frequency also increases.
Explanation:
The sound is perceived as louder if the amplitude increases, and softer if the amplitude decreases. ... The amplitude of a wave is related to the amount of energy it carries. A high amplitude wave carries a large amount of energy; a low amplitude wave carries a small amount of energy.
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Answer:
option A= Physical change
Explanation:
physical change:
" It is a change in which no new substance is formed"
Breaking of object like glass is the example of physical change because it is not change into another object. It effect the form of object but can not change the chemical composition.
Chemical changes:
" it is a change in which one substance is converted into new product through chemical reaction".
During the chemical changes the types and the number of atom remain same but their arrangement changed.
for example: burning of wood, baking of cake, digesting food, resting of iron etc.
Physical reaction:
" physical reaction occur during the molecular rearrangement. There is no chemical change occur"
In this type of changes no bonds are break to form new bonds, for example boiling point.
Chemical reaction:
" chemical reaction occur when molecules are chemically react with each others and bonds formation and breaking is also occur"
Answer:
C. CH3COOH, Ka = 1.8 E-5
Explanation:
analyzing the pKa of the given acids:
∴ pKa = - Log Ka
A. pKa = - Log (1.0 E-3 ) = 3
B. pKa = - Log (2.9 E-4) = 3.54
C. pKa = - Log (1.8 E-5) = 4.745
D. pKa = - Log (4.0 E-6) = 5.397
E. pKa = - Log (2.3 E-9) = 8.638
We choose the (C) acid since its pKa close to the expected pH.
⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):
- pH = pKa + Log ([CH3COO-]/[CH3COOH])
∴ pH = 4.5
∴ pKa = 4.745
⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])
⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])
⇒ 0.5692 = [CH3COO-]/[CH3COOH]
∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]
⇒ 1.8 E-5 = [H3O+](0.5692)
⇒ [H3O+] = 3.1623 E-5 M
⇒ pH = - Log ( 3.1623 E-5 ) = 4.5
