The population of big fish would increase because they have more food
<span>Na (sodium) is highly electropositive. Its has 1 electron in its outermost orbit which is transferred to an electronegative atom to form an ionic bond.
It only needs to get rid of one valence electron to take part in a reaction. That's how it's highly reactive.</span>
Explanation:
mass of an object affect the density and object if the the volume is keep constant.
here we can explain by using the example.
if the mass of an object is 30kg. and volume is 2m3
then density pf given substance is become 15kg/m3
again,
if the mass of the is 40kg/m3 and volume is same
then density is become 20kg/m3 (formula=kg/m3)
here above density and below density is becime different by changing the mass of an object and affect the density and objecy although volume is kept constant or same
Answer:
c. 0.2 M HNO₃ and 0.4 M NaF
.
Explanation:
A buffer is defined as the mixture of a weak acid with its conjugate base or a weak base with its conjugate acid.
A weak acid or weak base are defined as an acid or base that partially dissociates in aqueous solution. in contrast, a strong acid or base are acids or bases that is dissociated completely in water.
Thus:
a. 0,2M HNO₃ and 0.4 M NaNO₃. This is a mixture of a strong acid with its conjugate base. <em>IS NOT </em>a buffer.
b. 0.2 M HNO₃ and 0.4 M HF
. This is a mixture of two strong acids. <em>IS NOT </em>a buffer.
c. 0.2 M HNO₃ and 0.4 M NaF
. NaF is the conjugate base of a weak acid as HF is.
The reaction of HNO₃ with NaF is:
HNO₃ + NaF → HF + NaNO₃
That means that in solution you will have a weak acid (HF) with its conjugate base (NaF). Thus, this mixture <em>IS </em>a buffer.
d. 0.2 M HNO₃ and 0.4 M NaOH. This is the mixture of a strong acid with a strong base, thus, this <em>IS NOT </em>a buffer.
I hope it helps!
Answer:
2.5 atm
Explanation:
V1 = 28L
T1 = 45°C = (45 + 273.15)K = 318.15K
V2 = 34L
T2 = 35°C = (35 + 273.15)K = 308.15k
P2 = 2.0atm
P1 = ?
From general gas equation,
(P1 × V1) / T1 = (P2 × V2) / T2
P2 × V2 × T1 = P1 × V1 × T2
P1 = (P2 × V2 × T1) / (V1 × T2)
P1 = (2.0 × 34 × 318.15) / (28 × 308.15)
P1 = 21634.5 / 8628.2
P1 = 2.5 atm
The initial pressure of the gas is 2.5atm
