For every meter, the equivalent measurements is 1000 millimeters. Hence in the problem where the number of millimeters is given, we divide the number by 1000 to get the number of meters. The answer here is 0.01123 m.
Answer:
36.66%
Explanation:
Step 1: Given data
- Mass of the sample: 2.875 g
Step 2: Calculate the mass of salt
The mass of the sample is equal to the sum of the masses of the components.
m(sample) = m(iron) + m(sand) + m(salt)
m(salt) = m(sample) - m(iron) - m(sand)
m(salt) = 2.875 g - 0.660 g - 1.161 g
m(salt) = 1.054 g
Step 3: Calculate the percent of salt in the sample
We will use the following expression.
%(salt) = m(salt) / m(sample) × 100%
%(salt) = 1.054 g / 2.875 g × 100% = 36.66%
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Answer:
pH = 12.15
Explanation:
To determine the pH of the HCl and KOH mixture, we need to know that the reaction is a neutralization type.
HCl + KOH → H₂O + KCl
We need to determine the moles of each compound
M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl
M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH
The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl
HCl + KOH → H₂O + KCl
3 m 4 m -
1 m 3 m
As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.
1 mmol of KOH has 1 mmol of OH⁻, so the [OH⁻] will be 1 mmol / Tot volume
[OH⁻] 1 mmol / 70 mL = 0.014285 M
- log [OH⁻] = 1.85 → pH = 14 - pOH → 14 - 1.85 = 12.15
Answer:
See attachment.
Explanation:
In the first step, a cyclic structure with a positive bromine is formed. The bromine shares the positive charge with the two carbons that it is bonded to, so the carbons are partially positive.
The second bromine atom then attacks the carbon center, coming in from below the first bromine atom ("backside attack") where the antibonding orbital of the second bromine atom is.
The stereochemistry of the mechanism causes the final product to be an anti-dibromocyclohexane.
