Answer:
6.52 × 10^14 Hz
i don't know if that's right tbh
Answer:
C.106 ml
Explanation:
Boyle's Law P₁V₁ = P₂V₂ , but we need to make sure our pressure is in the same units, in this case we will convert torr to atm
811 torr * (1atm/760torr) = 1.067105263 atm
P₁V₁ = P₂V₂
(1.067105263 atm)(11.1 mL) = (0.112 atm)(V₂)
V₂ = 11.84486842/0.112
V₂ = 105.7577538 mL = 106 mL
Nitrogen gained 4 electrons.
Because Nitrogen's redox number went from +6 to +2, it must have gained 4 electrons (-4) in order to achieve this number. Thus, Nitrogen is reduced.
Compete Question:
What mass of CDP (403 g mol−1) is in 10 mL of the buffered solution at the beginning of Experiment 1?
Passage: "16 mmol of CDP in 1 L of buffer"
Answer:
6.4 × 10-2 g
Explanation:

we are given from the question that 16 mmol of CDP is in 1 L of buffer
this mean that we have
moles of CDP in 1 liter of buffer.
so the mass of CDP in one liter of buffer will be calculate as,
mass of CDP =
× 403g mol−1
=
= 6.4 g/L
But because the question
asks us about the mass of CDP in 10 mL of solution, we will go further to calculate it like this:
6.4 g/L × 10 mL
6.4 g/L × 0.01 L =
Answer:
Explanation:
A, D,E and B i think is the correct answers