Answer:
0.0847M is molarity of sodium hydrogen citrate in the solution
Explanation:
The 2.0%(w/v) solution of sodium hydrogen citrate contains 2g of the solute in 100mL of solution. To find the molarity of the solution we need to convert the mass of solute to moles using molar mass and the mL of solution to Liters because molarity is the ratio between moles of sodium hydrogen citrate and liters of solution.
<em>Moles Na2C6H6O7:</em>
<em>Molar Mass:</em>
2Na: 2*22.99g/mol: 45.98g/mol
6C: 6*12.01g/mol: 72.01g/mol
6H: 6*1.008g/mol: 6.048g/mol
7O: 7*16g/mol: 112g/mol
45.98g/mol + 72.01g/mol + 6.048g/mol + 112g/mol = 236.038g/mol
Moles of 2g:
2g * (1mol / 236.038g) = <em>8.473x10⁻³ moles</em>
<em />
<em>Liters solution:</em>
100mL * (1L / 1000mL) = <em>0.100L</em>
<em>Molarity:</em>
8.473x10⁻³ moles / 0.100L =
<h3>0.0847M is molarity of sodium hydrogen citrate in the solution</h3>
Answer:
Explanation:
2 NO2(g) ⇄ N2O4(g)
Adding Argon to this reaction will have NO effect. Catalysts nor inert gases have an affect on equilibrium conditions.
Only changes in concentration, temperature conditions and pressure-volume conditions (unless both sides have equal molar volumes) will affect the equilibria.
NH4OH(aq) ⇄ NH3(g) + H2O(l)
Removing ammonia from reaction equilibrium causes the reaction to shift right to replace removed ammonia. => Think of the reaction as being on a seesaw => removing ammonia from the product side tilts the seesaw left causing the NH₄OH to decompose and deliver more NH₃ and H₂O to the product side to increase weight on that side and level the seesaw. :-)
We observe that heat capacity of salted water we will find that it is less than pure water. We now that it takes less energy to increase the temperature of the salt water 1°C than pure water. Which means that the salted water heats up faster and eventually reaches to its boiling point first.
hope it helps
Answer:
225 mL of water must be added.
Explanation:
First we <u>calculate how many HCl moles are there in 516 mL of a 0.191 M solution</u>:
- 516 mL * 0.191 M = 98.556 mmol HCl
Now we use that number of moles (that remain constant during the <em>dilution process</em>) to <u>calculate the final volume of the 0.133 M solution</u>:
- 98.556 mmol / 0.133 M = 741 mL
We can <u>calculate the volume of water required</u> from the volume difference:
Answer:
Option A.
2Na + 2H2O —> 2NaOH + H2
Explanation:
To know which option is correct, we shall do a head count of the number of atoms present on both side to see which of them is balanced. This is illustrated below below:
For Option A:
2Na + 2H2O —> 2NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 2 Na
4 H >>>>>>>>>>>> 4 H
2 O >>>>>>>>>>>> 2 O
Thus, the above equation is balanced.
For Option B:
2Na + 2H2O —> NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 1 Na
4 H >>>>>>>>>>>> 3 H
2 O >>>>>>>>>>>> 1 O
Thus, the above equation is not balanced.
For Option C:
2Na + H2O —> 2NaOH + H2
Reactant >>>>>>> Product
2 Na >>>>>>>>>>> 2 Na
2 H >>>>>>>>>>>> 4 H
1 O >>>>>>>>>>>> 2 O
Thus, the above equation is not balanced.
For Option D:
Na + 2H2O —> NaOH + 2H2
Reactant >>>>>>> Product
1 Na >>>>>>>>>>> 1 Na
4 H >>>>>>>>>>>> 5 H
2 O >>>>>>>>>>>> 1 O
Thus, the above equation is not balanced.
From the illustrations made above, only option A is balanced.
