ok so this was really confusing, but i think i know the answer. D. ...hairy and ancient hermit crabs and swift, darting minnows and sometimes a
crumbling sand dollar.
Answer:
3.336.
Explanation:
<em>Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).</em>
<em />
So, <em>concentration of excess acid = [(NV)acid - (NV)base]/V total</em> = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = <em>1.18 x 10⁻³ M.</em>
<em></em>
<em> For weak acids; [H⁺] = √Ka.C</em> = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = <em>4.61 x 10⁻⁴ M.</em>
∵ pH = - log[H⁺].
<em>∴ pH = - log(4.61 x 10⁻⁴) = 3.336.</em>
Answer:
No, it is not sufficient
Please find the workings below
Explanation:
Using E = hf
Where;
E = energy of a photon (J)
h = Planck's constant (6.626 × 10^-34 J/s)
f = frequency
However, λ = v/f
f = v/λ
Where; λ = wavelength of light = 325nm = 325 × 10^-9m
v = speed of light (3 × 10^8 m/s)
Hence, E = hv/λ
E = 6.626 × 10^-34 × 3 × 10^8 ÷ 325 × 10^-9
E = 19.878 × 10^-26 ÷ 325 × 10^-9
E = 19.878/325 × 10^ (-26+9)
E = 0.061 × 10^-17
E = 6.1 × 10^-19J
Next, we work out the energy required to dissociate 1 mole of N=N. Since the bond energy is 418 kJ/mol.
E = 418 × 10³ ÷ 6.022 × 10^23
E = 69.412 × 10^(3-23)
E = 69.412 × 10^-20
E = 6.9412 × 10^-19J
6.9412 × 10^-19J is required to break one mole of N=N bond.
Based on the workings above, the photon, which has an energy of 6.1 × 10^-19J is not sufficient to break a N=N bond that has an energy of 6.9412 × 10^-19J
Density is calculated as mass divided by volume. If we are given an ice cube of side length 8.00 cm, then the volume of the cube is equivalent to (8.00 cm)^3 = 512 cm^3. Since we have a given mass of 476 g, we can divide:
476 g / 512 cm^3 = 0.930 g/cm^3
So the density of the sample of ice is 0.930 g/cm^3.
