In a single molecule of water two hydrogen atoms are bonded to a single oxygen atom by

Suppose that an element has two isotopes. The average atomic mass of the element is 45.737 u . One isotope has a mass of 40.149

Ugo [173]

Answer:

The answer to your question is 49.08 u

Explanation:

Data

Average atomic mass = 45.737 u

Isotope 1 = 40.149 u abundance = 37.46%

Isotope 2 = ?

Process

1.- Calculate the abundance of isotope 2

abundance isotope 1 + abundance isotope 2 = 100%

37.46 + abundance isotope 2 = 100

abundance isotope 2 = 100 - 37.46

abundance isotope 2 = 62.54

2.- Calculate the mass of isotope 2

Average atomic mass = (abundance x mass isotope 1) + (abundance x

mass isotope 2)

45.737 = (0.3746 x 40.149) + (62.54 x mass isotope 2)

mass isotope 2 = [45.737 - (0.3746 x 40.149)]/ 62.54

mass isotope 2 = [45.737 - 15.0398]/ 62.54

mass isotope 2 = 30.6972/62.54

mass isotope 2 = 0.4908 x 100

mass isotope 2 = 49.08 u

7 0

3 years ago

odium carbonate (Na2CO3Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be add

Arte-miy333 [17]

Answer : The mass of sodium carbonate added to neutralize must be, $6.54\times 10^3kg$

Explanation :

First we have to calculate the moles of $H_2SO_4$ .

$\text{Moles of }H_2SO_4=\frac{\text{Mass of }H_2SO_4}{\text{Molar mass of }H_2SO_4}$

Given:

Molar mass of $H_2SO_4$ = 98 g/mole

Mass of $H_2SO_4$ = $6.05\times 10^3kg=6.05\times 10^6g$

Conversion used : (1 kg = 1000 g)

Now put all the given values in the above expression, we get:

$\text{Moles of }H_2SO_4=\frac{6.05\times 10^6g}{98g/mol}=6.17\times 10^4mol$

The moles of $H_2SO_4$ is, $6.17\times 10^4mol$

Now we have to calculate the moles of $Na_2CO_3$

The balanced neutralization reaction is:

$Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+H_2CO_3$

From the balanced chemical reaction we conclude that,

As, 1 mole of $H_2SO_4$ neutralizes 1 mole of $Na_2CO_3$

So, $6.17\times 10^4mol$ of $H_2SO_4$ neutralizes

Now we have to calculate the mass of $Na_2CO_3$

$\text{ Mass of }Na_2CO_3=\text{ Moles of }Na_2CO_3\times \text{ Molar mass of }Na_2CO_3$

Molar mass of $Na_2CO_3$ = 106 g/mole

$\text{ Mass of }Na_2CO_3=(6.17\times 10^4mol)\times (106g/mole)=6.54\times 10^6g=6.54\times 10^3kg$

Thus, the mass of sodium carbonate added to neutralize must be, $6.54\times 10^3kg$

3 0

4 years ago

(d) What is the mass percentage of water in CdCl2.4H2O?

Zielflug [23.3K]

The mass percentage of water (H₂O) in cadmium chloride tetra hydratesolution is equal to the 28.2%.

<h3>How do we calculate mass percentage?</h3>

Mass percentage of any substabce present in any solution will be calculated as:
Mass percent = (Mass of substance/Total mass of solution)×100%

According to the question,

Mass of water (H₂O) = 18.02 g/mol

Mass of solvent (CdCl₂.4H₂O) = 183.32 + 4(18.02) = 255.4 g/mol

On putting values, we get

Mass percent = (18.02 / 255.4) × 100% = 28.2%

Hence mass percent of water is 28.2%.

To know more about mass percent, visit the below link:

brainly.com/question/13896694

6 0

2 years ago

Read 2 more answers

What volume of 1.00 m hcl in liters is needed to react completely (with nothing left over) with 0.750 l of 0.100 m na2co3?

kotykmax [81]

The balanced equation for the reaction is as follows
Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O
stoichiometry of Na₂CO₃ to HCl is 1:2
number of Na₂CO₃ moles reacted = molarity x volume
number of Na₂CO₃ moles = 0.100 mol/L x 0.750 L = 0.0750 mol
according to molar ratio of 1:2
1 mol of Na₂CO₃ reacts with 2 mol of HCl
then 0.0750 mol of Na₂CO₃ mol reacts with - 2 x 0.0750 = 0.150 mol
molarity of given HCl solution is 1.00 mol/L
molarity is defined as the number of moles of solute in 1 L of solution
there are 1.00 mol in 1 L of solution
therefore there are 0.150 mol in - 0.150 mol / 1.00 mol/L = 0.150 L
volume of HCl required is 0.150 L

3 0

3 years ago

Identify the oxidizing agent and the reducing agent for 4Li(s) + O_2 (g) to 2Li_2O(s).

Doss [256]

The reaction is:

<span>4Li(s) + O2 (g) = 2Li+ + 2O-2(s).

The oxidizing agent is the one that is being reduced which is oxygen where the charge changed from neutral to -2 while the reducing agent is the on being oxidized which is lithium where the charge change from neutral to +1.</span>

7 0

3 years ago