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ale4655 [162]
3 years ago
12

If one mole of PtCl2·3NH3 reacts with AgNO3 to produce one mole of AgCl(s), how can the formula PtCl2·3NH3 be re-written to show

the proper coordination sphere?
a) [Pt(NH3)3Cl2]
b) [Pt(NH3)3Cl]Cl
c) [Pt(NH3)3Cl3]-
d) [Pt(NH3)3]Cl2
Chemistry
1 answer:
Tresset [83]3 years ago
4 0
The correct answer is b) [Pt(NH₃)₃Cl]Cl
Due to:
Chloride ion which is present outside the co-ordination sphere will react with AgNO₃ & since the reaction produces only one mole of AgCl so one Cl only outside the sphere and the formula is: [Pt(NH₃)₃Cl]Cl<span />
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Which of these is a reason why nuclear power is important?
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The concentration of an additive in a standard sample of palm oil was measured 6 times and the following results were obtained:
Lerok [7]

In the data, 0.20 ppm is an outlier and this can be rejected if there is a 95% confidence level.

<h3>What is an outlier?</h3>

When analyzing data an outlier is a value that is abnormal or too different from other data. In the case presented 0.20 can be tagged as an outlier because other values such as 0.11, 0.12, 0.13, and 0.14 are similar while 0.20 is outside this range.

<h3>Should this piece of data be rejected?</h3>

The general rule is that if there is a 95% of confidence or higher you can reject an outlier, knowing the other data occurs 95% of the time, and therefore the outlier is improbable.

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7 0
2 years ago
Starting with 2.50 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0C, a chemist first heats the gas at con
arsen [322]

Answer:

a)  T_b=590.775k

b)  W_t=1.08*10^4J

d)  Q=3.778*10^4J

d)  \triangle V=4.058*10^4J

Explanation:

From the question we are told that:

Moles of N2 n=2.50

Atmospheric pressure P=100atm

Temperature t=20 \textdegree C

                      t = 20+273

                     t = 293k

Initial heat Q=1.36 * 10^4 J

a)

Generally the equation for change in temperature is mathematically given by

\triangle T=\frac{Q}{N*C_v}

  Where

  C_v=Heat\ Capacity \approx 20.76 J/mol/K

T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }

T_b-293k=297.775

T_b=590.775k

b)

Generally the equation for ideal gas is mathematically given by

 PV=nRT

For v double

 T_c=2*590.775k

 T_c=1181.55k

Therefore

PV=Wbc

Wbc=(2.20)(8.314)(1181_590.778)

Wbc=10805.7J

Total Work-done W_t

W_t=Wab+Wbc

W_t=0+1.08*10^4

W_t=1.08*10^4J

c)

Generally the equation for amount of heat added is mathematically given by

Q=nC_p\triangle T

Q=2.20*2907*(1181.55-590.775)\\

Q=3.778*10^4J

d)

Generally the equation for change in internal energy of the gas is mathematically given by

\triangle V=nC_v \triangle T

\triangle V=2.20*20.76*(1181.55-293)k

\triangle V=4.058*10^4J

3 0
3 years ago
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