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steposvetlana [31]
2 years ago
7

explain how to determine the number of atoms, cations, and anions in ionic compounds. use an example to explain. ​

Chemistry
1 answer:
Damm [24]2 years ago
8 0

Answer:

use coefficients and subscripts to determine how many atoms are in a compound. If there is no subscript or coefficient, assume it is 1. If there is a coefficient, multiply it with the subscripts. For counting cations and anions, determine first which is the anion and cation (anion = nonmetal, cation = metal), then count the number of that ion.

Example:

NaCl

one atom of Na, one atom of Cl. Since Na is a metal, it is a cation. Cl is a nonmetal, so it is an anion.

2CaCl2

2 atoms of Ca, 4 atoms of Cl. There are 2 cations, since Na is a metal, and 4 anions since Cl is a nonmetal

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If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi
zlopas [31]

Answer:

T_{eq}=28.9\°C

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

Q_{Cd}+Q_{W}=0

Thus, we insert mass, specific heat and temperatures to obtain:

m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}

Now, we plug in to obtain:

T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0
2 years ago
Silicon has 2 isotopes, Silicon-28 with an abundance of 90% and Silicon-30 with an abundance of 10%. Find the AAM for Silicon.
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Answer:

RAM

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3 years ago
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How many moles of sodium cyanide (NaCN) would be needed to produce 4.2 moles of sodium sulfate (Na2SO4)? H2SO4 + 2NaCN → 2HCN +
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To produce 4.2 moles of sodium sulfate, two times this amount of NaCN is required. This means that you would need 8.4 moles of sodium cyanide.

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Answer:

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