Answer:
142.82 g
Explanation:
The following data were obtained from the question:
Volume of water = 12 mL
Volume of water + gold = 19.4 mL
Density of gol= 19.3 g/cm³
Mass of gold =.?
Next, we shall determine the volume of the gold. This can be obtained as follow:
Volume of water = 12 mL
Volume of water + gold = 19.4 mL
Volume of gold =.?
Volume of gold = (Volume of water + gold) – (Volume of water)
Volume of gold = 19.4 – 12
Volume of gold = 7.4 mL
Finally, we shall determine the mass of the gold as follow:
Note: 1 mL is equivalent to 1 cm³
Volume of gold = 7.4 mL
Density of gol= 19.3 g/cm³ = 19.3 g/mL
Mass of gold =?
Density = mass /volume
19.3 = mass of gold /7.4
Cross multiply
Mass of gold = 19.3 × 7.4
Mass of gold = 142.82 g
Therefore, the mass of the gold pebble is 142.82 g
No, the water is not different it is just that the units of measurement in the US is different then those units in Europe. This is similar to how the US uses miles and Europe uses kilometers
2. B because it has a lower activation energy.
idk because you have no picture with the lines on it
Answer is: 6,16 kJ.
1) changing temperature of ice from -25°C to 0°C.
Q₁ = m·C·ΔT
Q₁ = 18 g · 2 J/g·°C · 25°C
Q₁ = 900 J.
m(H₂O) = 1mol · 18 g/mol = 18 g.
C - <span>specific heat of ice.
</span>2) changing temperature of water from 0°C to 70°C.
Q₁ = m·C·ΔT
Q₁ = 18 g · 4,18 J/g·°C · 70°C
Q₁ = 5266,8 J.
C - specific heat of water.
Q = Q₁ + Q₂ = 900 J + 5266,8 J
Q = 6166,8 J = 6,16 kJ.
Answer:
The ratio of acid to conjugate base is outside the buffer range of 10:1.
Explanation:
The Henderson-Hasselbalch equation for a buffer is
![\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D)
A buffer should have
![\dfrac{1}{10} \leq \dfrac{\text{[A$^{-}]$}}{\text{[HA]}} \leq \dfrac{10}{1}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B10%7D%20%5Cleq%20%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%5D%24%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%5Cleq%20%5Cdfrac%7B10%7D%7B1%7D)
For a solution that is 1.3 mol·L⁻¹ in HF and 1.3 mmol·L⁻¹ in KF, the ratio is
The ratio of acid to conjugate base is 1000:1, which is outside the range of 10:1.
A is wrong. NF is a weak acid.
C is wrong. The two species are a conjugate acid-base pair.
D is wrong. Salts of Group 1 metals are soluble.
