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2 years ago

What is the correct formula for combining Mg2+ and Br1+

Chemistry

1 answer:

True [87]2 years ago

6 0

Answer: The correct formula is $MgBr_2$

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here magnesium is having an oxidation state of +2 called as $Mg^{2+}$ cation and bromine $Br^{-}$ is an anion with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $MgBr_2$ .

The cations and anions being oppositely charged attract each other through strong coloumbic forces and form an ionic bond.

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What are some of the main ways to add energy to a reaction? plzzz help

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Answer:

Heat.

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Explanation:

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2 years ago

If some salt were accidently spilled as it was transferred from the balance to the cup would your calculated enthalpy of solutio

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Explanation:

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2 years ago

9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8

MissTica

Answer: 2.24 grams of Pb

Explanation:

Step 1

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

Step 2

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

Step 3

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

Step 4

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles * 1 ) 1

Moles of Pb =0.0108 moles

Step 5

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

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2 years ago

Which sentences in the passage are true?

user100 [1]

Answer:

First and last sentence .‍♀️

Explanation:

An important thing to keep in mind about the Reading Comprehension section of the GRE as we use PowerPrep online to study is that it is just that—reading comprehension. In other words, as difficult as it may seem, and it can be pretty tricky, the test makers will always give us all the information we need in the passage to answer the question.

7 0

3 years ago

Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:

dem82 [27]

Answer:

The percent yield of the reaction is 62.05 %

Explanation:

Step 1: Data given

Volume of methane = 25.5 L

Pressure of methane = 732 torr

Temperature = 25.0 °C = 298 K

Volume of water vapor = 22.0 L

Pressure of H2O = 704 torr

Temperature = 125 °C

The reaction produces 26.0 L of hydrogen gas measured at STP

Step 2: The balanced equation

CH4(g) + H2O(g) → CO(g) + 3H2(g)

Step 3: Calculate moles methane

p*V = n*R*T

⇒with p = the pressure of methane = 0.963158 atm

⇒with V = the volume of methane = 25.5 L

⇒with n = the moles of methane = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 298 K

n = (p*V) / (R*T)

n = (0.963158 * 25.5 ) / ( 0.08206 * 298)

n = 1.0044 moles

Step 4: Calculate moles H2O

p*V = n*R*T

⇒with p = the pressure of methane = 0.926316 atm

⇒with V = the volume of methane = 22.0 L

⇒with n = the moles of methane = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 398 K

n = (p*V) / (R*T)

n = (0.926316 * 22.0) / (0.08206 * 398)

n = 0.624 moles

Step 5: Calculate the limiting reactant

For 1 mol methane we need 1 mol H2O to produce 1 mol CO and 3 moles H2

H2O is the limiting reactant. It will completely be consumed (0.624 moles).

Methane is in excess. There will react 0.624 moles. There will remain 1.0044 - 0.624 moles = 0.3804 moles methane

Step 6: Calculate moles hydrogen gas

For 1 mol methane we need 1 mol H2O to produce 1 mol CO and 3 moles H2

For 0.624 moles H2O we'll have 3*0.624 = 1.872 moles

Step 9: Calculate volume of H2 at STP

1.0 mol at STP has a volume of 22.4 L

1.872 moles has a volume of 1.872 * 22.4 = 41.9 L

Step 10: Calculate the percent yield of the reaction

% yield = (actual yield / theoretical yield) * 100 %

% yield = ( 26.0 L / 41.9 L) *100 %

% yield = 62.05 %

The percent yield of the reaction is 62.05 %

6 0

2 years ago

Read 2 more answers