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3 years ago

Examine the following electron configuration:

Chemistry

1 answer:

maksim [4K]3 years ago

5 0

Answer:

The answer to 12 is Indium... you should be able to figure out the rest from there with help from the internet since i've given you the element

Explanation:

I would usually do it but I got a ton of hw

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Ni(NO3)2 (aq) + 2NaCl (aq)------>________________________ + ________________________

Alex787 [66]

Explanation:

Nickel(II) nitrate = Ni(NO3)2

Sodium Chloride = 2NaCl

Sodium Nitrate = 2NaNO3

Nickel(II) Chloride = NiCl2

Balanced equation :

Ni(NO3)2 + 2NaCl → 2NaNO3 + NiCl2

Complete Ionic equation:

Ni2+(aq) + 2NO3-(aq) + 2Na+(aq) + 2Cl-(aq) = 2Na+(aq) + 2NO3-(aq) + Ni2+(aq) + 2Cl-(aq)

Net ionic equation:

The balanced net ionic equation includes only species that are involved in the reaction. It can be found by removing ions that occur on both the reactant and product side of the complete ionic equation.

All the ions are spectator ions so there is no reaction.

4 0

2 years ago

Rick is creating a love potion for Morty. To make the potion, Rick's needs 51 mL of a mixture solution where 40% is carbonated w

Deffense [45]

Answer:

The amount of the first solution rick needs to mix together to create the love portion is 8.5 mL

Explanation:

So as to make the love potion, we have;

The percentage of carbonated water in the love portion = 40%

The percentage of green tea in the first solution = 65%

The percentage of carbonated water in the first solution = 15%

The percentage of whole milk in the first solution = 20%

The percentage of orange juice in the second solution = 17%

The percentage of lemonade in the second solution = 38%

The percentage of carbonated water in the second solution = 45%

Let 'x' represent the volume in mL of the first solution added to make the love portion, and let 'y' be the volume in mL of the second solution added to make the love portion, we have;

x + y = 51...(1)

0.15·x + 0.45·y = 0.40×51 = 20.4

0.15·x + 0.45·y = 20.4...(2)

Solving the system of simultaneous equation by making 'y' the subject of each of the equation gives;

For equation (1)

y = 51 - x

For equation (2)

y = 20.4/0.45 - (0.15/0.45)·x = 136 - 3·x

y = 136/3 - (1/3)·x

Equating the two equations of 'y', gives;

51 - x = 136/3 - (1/3)·x

51 - 136/3 = x - (1/3)·x

17/3 = (2/3)·x

(2/3)·x = 17/3

x = (3/2) × (17/3) = 17/2 = 8.5

x = 8.5

y = 51 - x = 42.5

y = 42.5

Therefore, the amount of the first solution rick needs to mix together to create the love portion, x = 8.5 mL

8 0

3 years ago

Using the following general equation for a nuclear reaction, complete the statements below.

sammy [17]

Atomic number is written in subscript:
Uranium atomic number =92

Mass of neutron is the superscript value times the quantity (moles):
3*1= 3

Atomic mass is approximately equivalent to the number of protons and neutrons in the atom (the mass number). It's written in superscript:

Kr=92

No. of neutrons = Mass number (in superscript) - Atomic number ( in subscript)

= 141-56= 85

No. of neutrons in reactants are: 3

8 0

4 years ago

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An igneous rock is composed primarily of quartz and feldspar. what type of magma most likely formed this rock?

Zarrin [17]

Answer:

4. Felsic

Explanation:

This is a felsic magma.

A felsic magma has high content of silicate minerals. Quartz and feldspar are silicate minerals with a high content of silica in them. Magma with such a rich composition of quartz and feldspar is known as a felsic magma.

A mafic magma is one with low silica content.

8 0

3 years ago

Read 2 more answers

If 252 grams of iron are reacted with 321 grams of chlorine gas, what is the mass of the excess reactant leftover after the reac

mario62 [17]

Answer:

Iron is in excess.

1) The mass of the iron remaining = 83.38 grams

2) Ethane is in excess. There will remain 90.06 grams ethane

Explanation:

Step 1: Data given

Mass of iron = 252 grams

Mass of Cl2 = 321 grams

Molar mass of Fe = 55.845

Molar mass of Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Fe(s)+3Cl2(g)⟶2FeCl3(s)

Step 3: Calculate moles

Moles = mass / molar mass

Moles Fe = 252.0 grams / 55.845 g/mol = 4.512 moles

Moles Cl2 = 321.0 grams / 70.90 g/mol = 4.528 moles

Step 4: Calculate the limiting reactant

For 2 moles Fe we need 3 moles Cl2 to produce 2 moles Fecl3

Cl2 is the limiting reactant. It will completely be consumed (4.528 moles).

Fe is in excess. There will 4.528 * 2/3 = 3.019 moles be consumed

There will remain 4.512 - 3.019 = 1.493 moles of Fe

The mass of the iron remaining = 1.493 * 55.845 g/mol =83.38 grams

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If 152 grams of ethane (C2H6) are reacted with 231 grams of oxygen gas, what is the excess reactant?

Step 1: Data given

Mass of ethane = 152.0 grams

mass of O2 =231.0 grams

Molar mass of ethane = 30.07 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

2C2H6(g) + 7O2(g) ⟶ 4CO2(g) + 6H2O(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles ethane = 152.0 grams / 30.07 g/mol = 5.055 moles

Moles O2 = 231.0 grams / 32.0 g/mol = 7.22 moles

Step 4: Calculate limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed (7.22 moles).

Ethane is in excess. There will react 7.22 * 2/7 = 2.06 moles

There will remain 5.055 - 2.06 = 2.995 moles ethane

2.995 moles ethane = 2.995 * 30.07 g/mol = 90.06 grams ethane

8 0

4 years ago

Read 2 more answers