Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:
Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:
So that we are now able to calculate the density in g/mL first:
Now, we proceed to the conversion to lb/in³ by using the following setup:
Regards!
Answer:
<em>1 mole is equal to 1 moles NaOH, or 39.99711 grams.</em>
Explanation:
<em>Hope this helps have a nice day :)</em>
Answer:
I think this is because math and chemistry go together and the math problems are science related.
Explanation:
17.8 mL NaOH
<em>Step 1.</em> Write the chemical equation
Fe^(2+) + 2NaOH → Fe(OH)2 + 2Na^(+)
<em>Step 2.</em> Calculate the moles of Fe^(2+)
Moles of Fe^(2+) = 500 mL Fe^(2+) × [0.0230 mmol Fe^(2+)]/[1 mL Fe^(2+)]
= 11.50 mmol Fe^(2+)
<em>Step 3.</em> Calculate the moles of NaOH
Moles of NaOH = 11.50 mmol Fe^(2+) × [2 mmol NaOH]/[1 mmol Fe^(2+)]
= 23.00 mmol NaOH
<em>Step 4.</em> Calculate the volume of NaOH
Volume of NaOH = 23.00 mmol NaOH × (1 mL NaOH/1.29 mmol NaOH)
= 17.8 mL NaOH
Answer : The maximum amount of nickel(II) cyanide is 
Explanation :
The solubility equilibrium reaction will be:
Initial conc. 0.220 0
At eqm. (0.220+s) 2s
The expression for solubility constant for this reaction will be,
![K_{sp}=[Ni^{2+}][CN^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BNi%5E%7B2%2B%7D%5D%5BCN%5E-%5D%5E2)
Now put all the given values in this expression, we get:
Therefore, the maximum amount of nickel(II) cyanide is
