Convert 6.0cc to cm cubed

An element has two naturally occurring isotopes, X-85 with a mass of 84.9118 amu and a natural abundance of 72.17%, and X-87 wit

QveST [7]

Answer:

This element is Rubidium (Rb) and has an average atomic mass of 85.468 u

Explanation:

The average mass of an element is calculated by taking the average of the atomic masses of its stable isotopes.

The enitre atomic mass = 100 % or 1

⇒ this consists of X-85 with 72.17 % abundance with atomic massof 84.9118 g/mol

72.17 % = 0.7217

⇒ this consists of X-87 with 27.83 % abundance with atomic mass of 86.9092 g/mol

27.83 % = 0.2783

To calculate the mass of this isotope we use the following:

0.7271 * 84.9118 + 0.2783 * 86.9092 =85.468 g/mol

This element is Rubidium(Rb) and has an average atomic mass of 85.468 u

5 0

3 years ago

What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?

AlladinOne [14]

The question is incomplete, here is the complete question:

The given chemical reaction is:

$2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3$

What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?

<u>Answer:</u> The theoretical yield of copper is 44.48 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of aluminium = 12.6 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{12.6g}{27g/mol}=0.467mol$

The given chemical equation follows:

$2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3$

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of copper

So, 0.467 moles of aluminium will produce = $\frac{3}{2}\times 0.467=0.7005mol$ of copper

Now, calculating the mass of copper from equation 1, we get:

Molar mass of copper = 63.5 g/mol

Moles of copper = 0.7005 moles

Putting values in equation 1, we get:

$0.7005mol=\frac{\text{Mass of copper}}{63.5g/mol}\\\\\text{Mass of copper}=(0.7005mol\times 63.5g/mol)=44.48g$

Hence, the theoretical yield of copper is 44.48 grams

4 0

3 years ago

1.How many mL of 0.401 M HI are needed to dissolve 5.97 g of BaCO3?

garri49 [273]

Answer:

The answer to your question is:

1.- volume = 0.151 l or 151 ml

2.- 0.241 l or 241 ml of NaOH

Explanation:

1.-

Data

V = ? HI = 0.401 M

BaCO3 = 5.97 g

2HI(aq) + BaCO3(s) ⇒ BaI2(aq) + H2O(l) + CO2(g)

MW BaCO3 = 137 + 12 + 48 = 197 g

197 g of BaCO3 ----------------- 1 mol

5.97 g ----------------- x

x = (5.97 x 1) /197

x = 0.03 mol of BaCO3

2 moles of HI ---------------- 1 mol of BaCO3

x ---------------- 0.03 mol of BaCO3

x = (0.03 x 2) / 1

x = 0.060 mol of HI

Molarity = moles / volume

volume = moles / molarity

volume = 0.060 / 0.401

volume = 0.151 l or 151 ml

2.-

V = ? NaoH 0.757 M

Co⁺² Volume = 167 ml 0.548 M

CoSO4(aq) + 2NaOH(aq) ⇒ Co(OH)2(s) + Na2SO4(aq)

Moles of Co = Molarity x volume

Moles of Co = 0.548 x 0.167

Moles of Co = 0.092

1 mol of CoSO4 -------------- 2 moles of NaOH

0.092 moles --------------- x

x = (0.092 x 2) /1

x = 0.183 moles of NaOH

Volume of NaOH = moles / molarity

= 0.183 / 0.757

= 0.241 l or 241 ml of NaOH

6 0

3 years ago

Which action would shift this reaction away from solid calcium fluoride and toward the dissolved ions? which action would shift

ANTONII [103]

Hello!

The chemical reaction for the dissolving of calcium fluoride is the following:

CaF₂(s) ⇄ Ca⁺²(aq) + 2F⁻(aq)

In this reaction, and according to Le Chatelier's principle, the action that would shift this reaction away from solid calcium fluoride and towards the dissolved ions is the removing of fluoride ions.

Le Chatelier's principle states that in an equilibrium reaction, the system would shift in the opposite direction of the changes. If we remove fluoride ions from the system, it will shift towards the formation of more fluoride ions by dissolving more Calcium Fluoride to achieve equilibrium again.

Have a nice day!

6 0

3 years ago

Bond length is the distance between the centers of two bonded atoms. On the potential energy curve, the bond length is the inter

mariarad [96]

Answer:

152 pm

Explanation:

According to the question, we can estimate the bond length from the given values of the atomic radii. This now is the upper limit of the bond length for the molecule.

Since we have that;

Atomic radius of H= 37.0 pm

Atomic radius of Br = 115.0 pm

Bond length = Atomic radius of H + Atomic radius of Br

Bond length = 37.0 pm + 115.0 pm

Bond length = 152 pm

5 0

2 years ago