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Komok [63]
3 years ago
14

Consider the balanced chemical equation,

Chemistry
2 answers:
Gemiola [76]3 years ago
7 0

Answer: we started with 10 ions of Mg and 20 ions of F.

Explanation: The given balanced equation is:

MgCl_2(aq)+2AgF(aq)\rightarrow 2AgCl(s)+MgF_2(aq)

From this balanced equation, 2 formula units of AgCl are formed by 2 formula units of AgF and 1 formula unit of MgCl_2 .

From here, the ratio between AgCl and AgF is 2:2 that is 1:1 and the ratio between AgCl and MgCl_2 is 2:1 ,

Since the ratio between AgCl and AgF is 1:1, 20 formula units of AgCl will be formed by 20 formula units of AgF.

The ratio between AgCl and MgCl_2 is 2:1, so 20 formula units of AgCl will be formed by 10 formula units of MgCl_2 .

One formula unit of MgCl_2 contains one ion of Mg, so 10 formula units of MgCl_2 will contain 10 ions of Mg.

Similarly, 1 formula unit of AgF contains one ion of F, so 20 formula units of AgF will contain 20 ions of F.

So, we started with 10 ions of Mg and 20 ions of F to make 20 formula units of AgCl.

Montano1993 [528]3 years ago
4 0
<span>We started with twenty ions of Mg and twenty ions of F. Producing 20 units of AgCl requires twenty MgCl2 since it is the only source of magnesium. Similarly it requires twenty AgF since it is the only source of silver.</span>
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Answer:

C = 0.2349 J/ (g °C)

Explanation:

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Heat, H = 4.90kJ = 4900 J

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If the cylindrical pistons are 25.000 cm in diameter at 20.0 ∘c, what should be the minimum diameter of the cylinders at that te
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Refer to the diagram shown below.

The piston supports the same load W at both temperatures.
The ideal gas law is
pV=nRT
where
p = pressure
V = volume
n = moles
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State 1:
T₁ = 20 C = 20+273 = 293 K
d₁ = 25 cm piston diameter

State 2:
T₂ = 150 C = 423 K
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Because V, n, and R remain the same between the two temperatures, therefore
\frac{p_{1}}{T_{1}} = \frac{p_{2}}{T_{2}}

If the supported load is W kg, then
p_{1} =  \frac{W \, N}{ \frac{\pi}{4} d_{1}^{2}} = \frac{4W \, N}{\pi (0.25 \, m)^{2}} =  20.3718W \, Pa
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p_{2} =  \frac{4W}{\pi d_{2}^{2}} \, Pa

\frac{p_{1}}{p_{2}} =  \frac{20.3718 \pi d_{2}^{2}}{4} = 16 d_{2}^{2}

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The minimum piston diameter at 150 C is 20.8 cm.

Answer: 20.8 cm diameter

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