The reaction between elemental phosphorus p4(s) and o2(g) to make p4o10(s):

1 answer:

6 0

<span>The reaction between elemental phosphorus p4(s) and o2(g) is a synthesis type of reaction where two elements combine together to form a compound. Balancing the equation, we get P4(s) + 5 O2(g) = P4O10(s). Here the number of oxygen molecules are balanced on either side by adding 5 molecules of O2 on the product side.</span>

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The number of p atoms in 1.0 g of ba3(po4)2 is:

marta [7]

First, we determine the number of moles of barium phosphate. This is done using:

Moles = mass / Mr
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Now, we see from the formula of the compound that each mole of barium phosphate has 2 moles of phosphorus (P). Therefore, the moles of phosphorus are:
0.002 * 2 = 0.004

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0.004 * 6.02 x 10²³

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A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net

Margarita [4]

Answer:

Pb^2+(aq) + 2F-(aq) → PbF2(s)

Explanation:

Step 1: Data given

sodium fluoride = NaF

lead(II)nitrate Pb(NO3)2

Step 2: The unbalanced equation

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)

Step 3: Balancing the equation

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)

On the left side we have 2x NO3 (in Pb(NO3)2), on the right side we have 1x NO3 (in NaNO3). To balance the amount of NO3 we hvae to multiply NaNO3 on the right side by 2.

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2NaNO3(aq)

On the left side we have 1x Na (in NaF), on the right side we have 2x Na (in 2NaNO3). To balance the amount of Na we have to multiply NaF on the left side by 2. Now the equation is balanced.

2NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2NaNO3(aq)

Step 4: Calculate net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2Na+(aq) + 2F-(aq) + Pb^2+(aq) + 2NO3-(aq) → PbF2(s) + 2Na+(aq) + NO3-(aq)

Pb^2+(aq) + 2F-(aq) → PbF2(s)

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</span>∴ ΔTf = Kf X m = 1.86 X 1.40 = 2.604 oC

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