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inessss [21]
3 years ago
15

The noble-gas notation for tin (Sn) will contain the symbol Ar Kr Xe Rn

Chemistry
2 answers:
salantis [7]3 years ago
5 0

Answer:

2. Kr

Explanation:

KIM [24]3 years ago
3 0

Answer: Noble-gas notation of Sn contains Kr.

Explanation: Tin ( Sn)  is an element having atomic number 50.

Nearest noble gas to this element is Krypton which has an atomic number 36.

Electronic configuration or noble-gas notation for Sn is written as :

Sn=[Kr]4d^{10}5s^25p^2

As seen from above, Noble gas Krypton having symbol 'Kr' is coming in the electronic configuration for Tin.

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4. Consider the following data: Metal Mass (9 Cu Specilic Heat_Wg % Temperature, 0,900 0.285 these two metals are placed in cont
lana66690 [7]

The heat will flow from copper to aluminum because Cu is at higher temperature. The heat liberated is -7.60kJ

When two metals at different temperatures are kept in contact, heat flows from hotter metal to colder metal until thermal equilibrium is reached.

Here Copper is at a temperature of 60 degree Celsius and aluminum is at 40 degree Celsius. Thus, heat will flow from Cu to Al.

In order to calculate the amount of heat liberated following calculations are required.

m1=262 g

T1=87 oC

Cp=0.385 J/g oC

T2=11.8 oC

The heat liberated can be expressed as follows:

Q=mCp(T2-T1)

Q=262 g*0.385 J/goC(11.8-87)oC

Q=-7585 J

=-7.60kJ

To learn more about heat check the link below:

brainly.com/question/13439286

#SPJ4

4 0
1 year ago
A sample of 211 g of iron (III) bromide is reacted with
Alisiya [41]

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

<h3>Further explanation</h3>

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

  • FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714

  • Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

\tt n=\dfrac{186}{78.0452}=2.38

Coefficient ratio from the equation FeBr₃ :  Na₂S = 2 : 3, so mol ratio :

\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793

So  FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

\tt \dfrac{6}{3}\times 0.714=1.428

6 0
3 years ago
What is the formula for the compound by calcium ions and chloride ions?
dimaraw [331]
To find the ratio of the the combination for the ion, write the charge of the cation as the subscript for the anion, and the charge of the anion as the subscript of the cation. This will make the charges effectively cancel and you will be left with a neutral ionic compound. Remember, that an ionic compound is made up of a metal and a nonmetal.
For Ca2+ and Cl-, you will get the neutral compound to be CaCl₂.
8 0
4 years ago
Which statement is true of diffusion?
raketka [301]

Answer:

Molecules move from areas of high concentration to areas of low concentration.

4 0
3 years ago
How many cm in 725 km?
tigry1 [53]
1 km is equal to 100,000 cm
725km \times 100000cm \\  = 72500000 \: km .cm
and you need to divide it to 1 km so that you can get rid of the km.
= 72500000cm
5 0
3 years ago
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