The reaction between K₂SO₄(aq) and SrI₂(aq) produces KI(aq) and SrSO₄(s) as products.
The reaction is
K₂SO₄(aq) + SrI₂(aq) → KI(aq)+ SrSO₄(s)
To balance the equation both side of the reaction should have same number of atoms in each element.
Right hand side of the reaction has 1 K, 1 I, 1 Sr, 1 S and 4 O atoms while 2 K, 2 I, 1 Sr,1 S and 4 O present in left hand side of the reaction.
Hence, number of I atoms and number of K atoms are not balanced.
To balance the K atoms we should add 2 before KI. Then I atoms will be 2 at the right hand side.
Hence, the balanced reaction equation is
K₂SO₄(aq) + SrI₂(aq) → 2KI(aq)+ SrSO₄(s)
Answer:
- Empirical:

- Molecular:

Explanation:
Hello,
In this case, based on the information regarding the combustion, the moles of carbon turn out:

Moreover, the moles of hydrogen:

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

Which is hexane.
Best regards.
I'd say the correct answer is: Noodles rising and falling apart in boiling water.
Answer:
Pb(s) ---> Pb+2 + 2e- is the anode
Cu+2(aq) + 2e- ---> Cu(s) is the cathode
Answer:
Concentration of sodium carbonate in the solution before the addition of HCl is 0.004881 mol/L.
Explanation:

Molarity of HCl solution = 0.1174 M
Volume of HCl solution = 83.15 mL = 0.08315 L
Moles of HCl = n



According to reaction , 2 moles of HCl reacts with 1 mole of sodium carbonate.
Then 0.009762 mol of HCl will recat with:

Moles of Sodium carbonate = 0.004881 mol
Volume of the sodium carbonate containing solution taken = 1L
Concentration of sodium carbonate in the solution before the addition of HCl:
![[Na_2CO_3]=\frac{0.004881 mol}{1 L}=0.004881 mol/L](https://tex.z-dn.net/?f=%5BNa_2CO_3%5D%3D%5Cfrac%7B0.004881%20mol%7D%7B1%20L%7D%3D0.004881%20mol%2FL)