3. Given the reaction:

If expending 3500 kcal is equal to a loss of 1.0 lb, how many days will it take Charles to lose 5.0 lb? Express your answer to t

Trava [24]

Answer:

WAIT

good luck on whatever this is for my dude.

3 0

2 years ago

There are three isotopes of X element (X):

Nutka1998 [239]

Answer:

Average atomic mass = 17.5 amu.

Explanation:

Given data:

X-17 isotope = atomic mass17.2 amu, abundance:78.99%

X-18isotope = atomic mass 18.1 amu, abundance 10.00%

X-19isotope = atomic mass:19.1 amu, abundance: 11.01%

Average atomic mass of X = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100

Average atomic mass = (78.99×17.2)+(10.00×18.1) +(11.01+ 19.1) /100

Average atomic mass = 1358.628 + 181 +210.291 / 100

Average atomic mass = 1749.919 / 100

Average atomic mass = 17.5 amu.

8 0

2 years ago

Consider the following reaction between calcium oxide and carbon dioxide: CaO(s)+CO2(g)→CaCO3(s) A chemist allows 14.4 g of CaO

sweet-ann [11.9K]

Answer:

Theoretical yield =26.03 g

Percent yield = 87%

Limiting reactant = CaO

Explanation:

Given data:

Mass of CaO = 14.4 g

Mass of CO₂ = 13.8 g

Actual yield of CaCO₃ = 22.6 g

Theoretical yield = ?

Percent yield = ?

Limiting reactant = ?

Solution:

Chemical equation:

CaO + CO₂ → CaCO₃

Number of moles of CaO:

Number of moles = Mass /molar mass

Number of moles = 14.4 g / 56.1 g/mol

Number of moles = 0.26 mol

Number of moles of CO₂:

Number of moles = Mass /molar mass

Number of moles = 13.8 g / 44 g/mol

Number of moles = 0.31 mol

Now we will compare the moles of CO₂ and CaO with CaCO₃ .

CO₂ : CaCO₃

1 : 1

0.31 : 0.31

CaO : CaCO₃

1 : 1

0.26 : 0.26

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

Mass of CaCO₃: Theoretical yield

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ =0.26 mol × 100.1 g/mol

Mass of CaCO₃ = 26.03 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 22.6 g/ 26.03 g × 100

Percent yield = 0.87× 100

Percent yield = 87%

Limiting reactant:

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

7 0

3 years ago

Determine if the data are qualitative or quantitative.

jonny [76]

Hi there!
Zinc: Is qualitative
Chlorine: is quantitative
Gallium: is neither
Nitrogen: is quantitative
Aluminum: is quantitative
If you need an explanation, please let me know !
Hope this helps and have a good day :) !
~Angel

8 0

3 years ago

Read 2 more answers

18-19. If a sample for spectrophotometric analysis is placed in a 10-cm cell, the absorbance will be 10 times greater than the a

miss Akunina [59]

Answer:

As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.

Explanation:

Absorbance of light by a reagent of concentration c, is given as

A = εcl

A = Absorbance

ε = molar absorptivity

c = concentration of reagent.

l = length of light path or length of the solution the light passes through.

So, if all.other factors are held constant, If a sample for spectrophotometric analysis is placed in a 10-cm cell, the absorbance will be 10 times greater than the absorbance in a 1-cm cell.

But the reagent blank solution is called a blank solution because it lacks the given reagent. A blank solution does not contain detectable amounts of the reagent under consideration. That is, the concentration of reagent in the blank solution is 0.

Hence, the Absorbance is subsequently 0. And increasing or decreasing the path length of light will not change anything. As long as it is a blank solution of the reagent, the Absorbance will be 0 regardless of the path length.

Hope this Helps!!!

5 0

3 years ago