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2 years ago

I need help please <33

Chemistry

1 answer:

Aliun [14]2 years ago

4 0

Answer:

Option 3 is correct.

The atomic nucleus of each element has a unique number of protons. Therefore, the energy of the electron layers of the atoms of each element is unique.

If we have the energy released from each electron transfer between the layers of the atom, we can identify the element.

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Please help ASAP! I’m lost right now.

SCORPION-xisa [38]

1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1

Chromium is strange because it moves on to the 4s orbital instead of filling the 3d orbital with that last electron. Tricky.

Mark as brainliest if this helped! :)

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3 years ago

Can somebody in 8thgarde help me with Examining Molecules and compounds ?

MakcuM [25]

I might be able to. I am in 8th grade science.

5 0

3 years ago

What is the mass of the solid NH4Cl formed when 75.5 g of NH3 is mixed with an equal mass of HCl? What is the volume of the gas

Gekata [30.6K]

Answer : The volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

The mass of $NH_4Cl$ produced is, 110.7 grams.

Explanation :

The balanced chemical reaction will be:

$NH_3+HCl\rightarrow NH_4Cl$

First we have to calculate the moles of $NH_3$ and HCl

$\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}$

Molar mass of NH_3 = 17 g/mole

$\text{Moles of }NH_3=\frac{75.5g}{17g/mole}=4.44mole$

and,

$\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}$

Molar mass of HCl = 36.5 g/mole

$\text{Moles of }HCl=\frac{75.5g}{36.5g/mole}=2.07mole$

Now we have to calculate the limiting and excess reagent.

From the balanced reaction we conclude that

As, 1 mole of HCl react with 1 mole of $NH_3$

So, 2.07 mole of HCl react with 2.07 mole of $NH_3$

From this we conclude that, $NH_3$ is an excess reagent because the given moles are greater than the required moles and HCl is a limiting reagent and it limits the formation of product.

The remaining moles of HCl gas = 4.44 - 2.07 = 2.37 moles

Now we have to calculate the volume of the gas remaining.

Using ideal gas equation :

PV = nRT

where,

P = Pressure of gas = 752 mmHg = 0.989 atm (1 atm = 760 mmHg)

V = Volume of gas = ?

n = number of moles of gas = 2.37 moles

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = $14.0^oC=273+14.0=287K$

Putting values in above equation, we get:

$0.989atm\times V=2.37mole\times (0.0821L.atm/mol.K)\times 287K$

V = 56.5 L

Now we have to calculate the moles of $NH_4Cl$

As, 1 mole of HCl react with 1 mole of $NH_4Cl$

So, 2.07 mole of HCl react with 2.07 mole of $NH_4Cl$

Now we have to calculate the mass of $NH_4Cl$

$\text{ Mass of }NH_4Cl=\text{ Moles of }NH_4Cl\times \text{ Molar mass of }NH_4Cl$

Molar mass of $NH_4Cl$ = 53.5 g/mole

$\text{ Mass of }NH_4Cl=(2.07moles)\times (53.5g/mole)=110.7g$

Thus, the volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

The mass of $NH_4Cl$ produced is, 110.7 grams.

3 0

3 years ago

1. Sodium hydroxide reacts with carbon dioxide according to the equation: 2NaOH(s) + CO2(g) →

Leno4ka [110]

The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH

<h3>What is Limiting reagent ?</h3>

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

Given chemical equation in balanced form ;

2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).

According to the Chemical equation ;

The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH

If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.

But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.
6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react

As 80 g NaOH produces 106 g of Na₂CO₃.

Therefore 5 g NaoH will produce ;

106 / 80 x 5 = 6.625 g

Learn more about limiting reagent here ;

brainly.com/question/11848702

#SPJ1

5 0

2 years ago

If a reaction vessel initially contains 5 mol S and 10 mil O2, how many moles of S will be in the reaction vessel once the react

Alex_Xolod [135]

Answer:

Explanation:

2S + 3O₂ = 2SO₃

2moles 3 moles

2 moles of S react with 3 moles of O₂

5 moles of S will react with 3 x 5 / 2 moles of O₂

= 7.5 moles of O₂ .

O₂ remaining unreacted = 10 - 7.5 = 2.5 moles .

All the moles of S will exhausted in the reaction and 2.5 moles of oxygen will be left .

6 0

3 years ago