Answer:
We have to add 9.82 grams of calcium acetate
Explanation:
Step 1: Data given
Molarity of the calcium acetate solution = 0.207 M
Volume = 300 mL = 0.300 L
Molar mass calcium acetate = 158.17 g/mol
Step 2: Calculate moles calcium acetate
Moles calcium acetate = molarity * volume
Moles calcium acetate = 0.207 M * 0.300 L
Moles calcium acetate = 0.0621 moles
Step 3: Calculate mass calcium acetate
Mass calcium acetate = moles * molar mass
Mass calcium acetate = 0.0621 moles * 158.17 g/mol
Mass calcium acetate = 9.82 grams
We have to add 9.82 grams of calcium acetate
Answer:
Explanation: When solutions of potassium iodide and lead nitrate are combined?
The lead nitrate solution contains particles (ions) of lead, and the potassium iodide solution contains particles of iodide. When the solutions mix, the lead particles and iodide particles combine and create two new compounds, a yellow solid called lead iodide and a white solid called potassium nitrate. Chemical Equation Balancer Pb(NO3)2 + KI = KNO3 + PbI2. Potassium iodide and lead(II) nitrate are combined and undergo a double replacement reaction. Potassium iodide reacts with lead(II) nitrate and produces lead(II) iodide and potassium nitrate. Potassium nitrate is water soluble. The reaction is an example of a metathesis reaction, which involves the exchange of ions between the Pb(NO3)2 and KI. The Pb+2 ends up going after the I- resulting in the formation of PbI2, and the K+ ends up combining with the NO3- forming KNO3. NO3- All nitrates are soluble. ... (Many acid phosphates are soluble.)
Answer:
3AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄
The coefficients are 3, 1, 3, 1
Explanation:
From the question given above, the following data were:
Silver chloride reacts with sodium phosphate to yield sodium chloride and silver phosphate. This can be written as follow:
AgCl + Na₃PO₄ —> NaCl + Ag₃PO₄
The above equation can be balanced as follow:
AgCl + Na₃PO₄ —> NaCl + Ag₃PO₄
There are 3 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 3 in front of NaCl as shown below:
AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄
There are 3 atoms of Cl on the right side and 1 atom on the left. It can be balance by putting 3 in front of AgCl as shown below:
3AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄
Thus, the equation is balanced.
The coefficients are 3, 1, 3, 1
