B; Seawater mixes with freshwater so the water has intermediate salinity
Explanation:
In an estuary, seawater mixes with freshwater so the water has intermediate salinity. Estuaries are usually located in transitional environments.
- Estuary is the wide part of a river where it nears the sea.
- This is called a transitional zone.
- Water from continental rivers usually fresh are brought in close contact with ocean water that is salty.
- The water here is said to be brackish as it is intermediate between salt and seawater.
- Organisms living in such terrain must be be well adapted to changing salinity.
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The grams that would be produced from 7.70 g of butanoic acid and excess ethanol is 7.923grams
calculation
Step 1: write the chemical equation for the reaction
CH3CH2CH2COOH + CH3CH2OH → CH3CH2CH2COOCH2CH3 +H2O
step 2: find the moles of butanoic acid
moles= mass/ molar mass
= 7.70 g/ 88 g/mol=0.0875 moles
Step 3: use the mole ratio to determine the moles of ethyl butyrate
moles ratio of CH3CH2CH2COOH :CH3CH2CH2COOCH2CH3 is 1:1 therefore the moles of CH3CH2CH2COOCH2CH3 = 0.0875 x78/100=0.0683moles
step 4: find mass = moles x molar mass
= 0.0683 moles x116 g/mol=7.923grams
Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
Answer:
6.022×10²² molecules
Explanation:
Given data:
Volume of nitrogen = 224 L
Pressure = standard = 1 atm
Temperature = standard = 273 K
Number of molecules = ?
Solution:
PV = nRT
1 atm × 224 L = n × 0.0821 atm.L/mol.K × 273 K
224 atm.L = n ×22.41 atm.L/mol
n = 224 atm.L/22.41 atm.L/mol
n = 10 mol
1 mole contain 6.022×10²³ molecules
10 mol×6.022×10²³ molecules/ 1 mol
60.22×10²³ molecules
6.022×10²² molecules
