When naming an ionic compound, write the name of the cation, which is the metal first. Then, write the name of the anion, which is the nonmetal. However, you remove the last 2-3 letters and replace suffixes.
1. RbF --> Rubidium Fluoride
Change fluorine to fluoride
2. CuO --> Copper (II) Oxide
Change oxygen to oxide. Oxide has a charge of -2. Since no subscripts are written, it means they have the same opposite charge. So, we use Copper (II).
<span>3. (NH</span>₄<span>)</span>₂<span>C</span>₂<span>O</span>₄ ---> Ammonium Oxalate
NH₄ is ammonia, but we change it to ammonium for polyatomic ions.
Answer:
Four times the original amount if only one orange was used
Explanation:
We can assume that the oranges all have equal voltages. Connecting them in series will have an increasing effect on the voltage delivered. In our case, this will produce 4 times the voltage of the circuit when only one orange is used.
Whenever simple cells are connected in series, the voltages of the individual cells are added up to form the voltage of the whole circuit.
Let us assume that the voltage of each of the oranges is approximately 0.9 volts. The Voltage produced when the 4 oranges are joined in series is 0.9 + 0.9 + 0.9 + 0.9 = 3.6 volts
B Very Acidic llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll
Answer:
Approximately 
, assuming that this acid is monoprotic.
Explanation:
Assume that this acid is monoprotic. Let 
denote this acid.
.
Initial concentration of without any dissociation:
![[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}](https://tex.z-dn.net/?f=%5B%7B%5Crm%20HA%7D%5D%20%3D%200.730%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D)
.
After 
of that was dissociated, the concentration of both 
and 
(conjugate base of this acid) would become:
.
Concentration of in the solution after dissociation:
.
Let ![[{\rm HA}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20HA%7D%5D)
, ![[{\rm H}^{+}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20H%7D%5E%7B%2B%7D%5D)
, and ![[{\rm A}^{-}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20A%7D%5E%7B-%7D%5D)
denote the concentration (in 
or 
) of the corresponding species at equilibrium. Calculate the acid dissociation constant 
for , under the assumption that this acid is monoprotic:
![\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7DK_%7B%5Crm%20a%7D%20%26%3D%20%5Cfrac%7B%5B%7B%5Crm%20H%7D%5E%7B%2B%7D%5D%20%5Ccdot%20%5B%7B%5Crm%20A%7D%5E%7B-%7D%5D%7D%7B%5B%7B%5Crm%20HA%7D%5D%7D%20%5C%5C%20%26%3D%20%5Cfrac%7B%280.09125%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%29%20%5Ctimes%20%280.09125%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%29%7D%7B0.63875%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%7D%5C%5C%5B0.5em%5D%26%5Capprox%201.30%20%5Ctimes%2010%5E%7B-2%7D%20%5Cend%7Baligned%7D)
.
Answer:
V₂ ≈416.7 mL
Explanation:
This question asks us to find the volume, given another volume and 2 temperatures in Kelvin. Based on this information, we must be using Charles's Law and the formula. Remember, his law states the volume of a gas is proportional to the temperature.
where V₁ and V₂ are the first and second volumes, and T₁ and T₂ are the first and second temperature.
The balloon has a volume of 600 milliliters and a temperature of 360 K, but the temperature then drops to 250 K. So,
- V₁= 600 mL
- T₁= 360 K
- T₂= 250 K
Substitute the values into the formula.
- 600 mL /360 K = V₂ / 250 K
Since we are solving for the second volume when the temperature is 250 K, we have to isolate the variable V₂. It is being divided by 250 K. The inverse o division is multiplication, so we multiply both sides by 250 K.
- 250 K * 600 mL /360 K = V₂ / 250 K * 250 K
- 250 K * 600 mL/360 K = V₂
The units of Kelvin cancel, so we are left with the units of mL.
- 250 * 600 mL/360=V₂
- 416.666666667 mL= V₂
Let's round to the nearest tenth. The 6 in the hundredth place tells us to round to 6 to a 7.
The volume of the balloon at 250 K is approximately 416.7 milliliters.
