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a 25.0ml dsmplr og 0.105M HCI was titrated with 31.5ml of NaOH. What is the concentration of the NaOH I got 0.0834M am I correct HCI + NaOH -----> H20 + NaCl 25.0ml HCI x 1L/1000mlx0.105 mol HCI/1L HCI X 1 mol NaOH/ I mol HCI = 0.0834M
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Hoped That Helped You
Answer:
V CH4(g) = 190.6 L
Explanation:
assuming ideal gas:
∴ STP: T =298 K and P = 1 atm
∴ R = 0.082 atm.L/K.mol
∴ moles (n) = 7.80 mol CH4(g)
∴ Volume CH4(g) = ?
⇒ V = RTn/P
⇒ V CH4(g) = ((0.082 atm.L/K.mol)×(298 K)×(7.80 mol)) / (1 atm)
⇒ V CH4(g) = 190.6 L
The first diagram is showing carboxylic acid because carboxyl contains a Carbon atom as the central atom bonded with OH group, a double bonded Oxygen, hydrogen ofc, and a lone R group.
2. It is ester group
3. Correct its ether
4. It's the exact diagram for ethyl alcohol, so correct
WOOT WOOOT