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MAVERICK [17]
3 years ago
12

A sample of gas is measured to be 375 ml at 1.15 atm. The pressure

Chemistry
1 answer:
andre [41]3 years ago
6 0

Answer:

Direct method:

1.15x375/2.05

=21.03mL

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3Co2 (aq) 6NO3(aq) 6Na (aq) 2PO43–(aq) → Co3(PO4)2(s) 6Na (aq) 6NO3(aq) Identify the net ionic equation for this reaction.
Burka [1]

Answer : The net ionic equation will be,

3Co^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Co_3(PO_4)_2(s)

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

3Co^{2+}(aq)+6NO_3^-(aq)+6Na^+(aq)+2PO_4^{3-}(aq)\rightarrow Co_3(PO_4)_2(s)+6Na^+(aq)+6NO_3^-(aq)

In this equation, NO_3^-(aq)\text{ and }Na^+(aq) are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

Thus, the net ionic equation will be,

3Co^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Co_3(PO_4)_2(s)

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3 years ago
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kondor19780726 [428]

Answer:

.156   M

Explanation:

diluting 25 to 100 cuts the M to 1/4 of the original

1/4 * .625 = .15625

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