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marysya [2.9K]
3 years ago
13

Enter the chemical formula for the cation present in the aqueous solution of AgNO3.

Chemistry
1 answer:
elixir [45]3 years ago
7 0

Answer:

Cation Ag⁺

Explanation:

As a ionic salt, AgNO₃ dissociates in water as this:

AgNO₃  →   Ag⁺ + NO₃⁻

It is a completely soluble salt.

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Which two elements are in the same period?A. hydrogen (H) and neon (Ne)B. tin (Sn) and antimony (Sb)C. fluorine (F) and chlorine
ArbitrLikvidat [17]

Answer:

C. fluorine (F) and chlorine (Cl)

D. arsenic (As) and antimony (Sb)

Explanation:

In the periodic table , all the elements are arranged according to the atomic number ,

and the elements are placed in groups and periods ,

The elements with similar chemical and physical properties are placed in a common group .

The elements present in the same group have the same number of valence electrons in the valence shell  .

Hence , from the given options ,

fluorine (F) and chlorine (Cl)  belongs to group 17 with 7 valence electrons in the outermost shell .

arsenic (As) and antimony (Sb) belong to group 15 with 3 valence electrons in the outermost shell .

8 0
3 years ago
WILL GIVE BRAINLIEST!! Determine the volume of a container that holds 2.4 mol of gas at STP.
kykrilka [37]

Answer:

2.4 mol x 22.4 liter = 53.76 liters

1 mole

Explanation:

5 0
2 years ago
5. A fireplace fire heating a room on a cold day is an example of what type of heat
Inessa [10]

Answer:

I think the answer is C- radiation

Explanation:

If its not C its A Hope this helps forgive me if i'm wrong :/

8 0
2 years ago
Is BaSO3 a poly atomic covalent
earnstyle [38]

Answer:

no it is a ionic compound

5 0
2 years ago
54.56 g of water at 80.4 oC is added to a calorimeter that contains 47.24 g of water at 40 oC. If the final temperature of the s
fomenos

Answer:

49.5J/°C

Explanation:

The hot water lost some energy that is gained for cold water and the calorimeter.

The equation is:

Q(Hot water) = Q(Cold water) + Q(Calorimeter)

<em>Where:</em>

Q(Hot water) = S*m*ΔT = 4.184J/g°C*54.56g*(80.4°C-59.4°C) = 4794J

Q(Cold water) = S*m*ΔT = 4.184J/g°C*47.24g*(59.4°C-40°C) = 3834J

That means the heat gained by the calorimeter is

Q(Calorimeter) = 4794J - 3834J = 960J

The calorimeter constant is the heat gained per °C. The change in temperature of the calorimeter is:

59.4°C-40°C = 19.4°C

And calorimeter constant is:

960J/19.4°C =

<h3>49.5J/°C</h3>

<em />

7 0
2 years ago
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