Plz help me!! Will mark BRAINLIEST!

Assuming all volume measurements are made at the same temperature and pressure, how many liters of water vapor can be produced w

maria [59]

For the reaction 2 K + F2 --> 2 KF,
consider K atomic wt. = 39
23.5 g of K = 0.603 moles, hence following the molar ratio of the balanced equation, 0.603 moles of potassium will use 0.3015 moles of F2. (number of moles, n = 0.3015)

Now, following the ideal gas equation, PV = nRT
P = 0.98 atm
V = unknown
n = 0.3015 moles
R = 82.057 cm^3 atm K^-1mole^-1 (unit of R chosen to match the units of other parameters; see the reference below)
T = 298 K
Solving for V,
V = (nRT)/P = (0.3015 mol * 82.057 cm^3 atm K^-1 mol^-1 * 298 K)/(0.98 atm)
solve it to get 7517.6 cm^3 as the volume of F2 = 7.5176 liters of F2 gas is needed.

2. Use the formula: volume1 * concentration 1 = volume 2 * concentration 2
where, volume 1 and concentration 1 are for solution 1 and volume 2 and solution 2 for solution 2.

Solution 1 = 12.3 M NaOH solution
Solution 2 = 1.2 M NaOH solution

<span> Solving for volume 1, volume 1 = (12.4 L * 1.2 M)/12.3 M = 0.1366 L </span>

5 0

3 years ago

Determine the moles of H+ reacting with the metal based on the following experimental data. 10.00 mL of 1.00 M HCl solution is a

musickatia [10]

Answer:

Approximately $1.876 \times 10^{-3}\; \rm mol$ .

Explanation:

Convert both volumes to standard units (that is: liters.)

$10.00 \; \rm mL = 10.00 \times 10^{-3}\; \rm L = 1.000 \times 10^{-2}\; \rm L$ .
$27.08 \; \rm mL = 27.08 \times 10^{-3}\; \rm L = 2.708 \times 10^{-2}\; \rm L$ .

Number of moles of $\rm HCl$ initially present (in the $10.00\; \rm mL$ solution at $1.00\; \rm M$ .)

$n(\mathrm{HCl}, \, \text{initial}) = \displaystyle c \cdot V = 1.00\times 1.000 \times 10^{-2}= 1.000\times 10^{-2}\; \rm mol$ .

Number of moles of $\rm NaOH$ from the titration:

$n(\mathrm{NaOH}) = c \cdot V = 0.30 \times 2.708 \times 10^{-2} = 8.124 \times 10^{-3}\; \rm mol$ .

$\rm NaOH$ neutralizes $\rm HCl$ at a $1:1$ ratio:

$\rm HCl + NaOH \to NaCl + H_2O$ .

Hence, $n(\mathrm{HCl},\, \text{leftover}) = n(\mathrm{NaOH}) = 8.124 \times 10^{-3}\; \rm mol$ .

$\begin{aligned}&n(\mathrm{HCl},\, \text{consumed}) \\ =& n(\mathrm{HCl},\, \text{initial}) - n(\mathrm{HCl},\, \text{leftover}) \\ =& 1.000\times 10^{-2} - 8.124\times 10^{-3} \\ =& 1.876 \times 10^{-3}\; \rm mol\end{aligned}$ .