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PtichkaEL [24]
2 years ago
13

A student is given two 10g samples, each a mixture of only nacl(s) and kcl(s) but in different proportions. Which of the followi

ng pieces of information could be used to determine which mixture has the higher proportion of KCl(s) ?
A. The volume of each mixture
B. The mass of Cl in each mixture
C. The number of isotopes of Na and K
D. The reaction of each mixture with water
Chemistry
1 answer:
earnstyle [38]2 years ago
4 0

Answer:

a

Explanation:

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4. DBearded waste of Co-60 must be stored until it is no longer radioactive. Cobalt-60
Bingel [31]

464 g radioisotope was present when the sample was put in storage

<h3>Further explanation</h3>

Given

Sample waste of Co-60 = 14.5 g

26.5 years in storage

Required

Initial sample

Solution

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}

t = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

Half-life of Co-60 = 5.3 years

Input the value :

\tt 14.5=No.\dfrac{1}{2}^{26.5/5.3}\\\\14.5=No.\dfrac{1}{2}^5\\\\No=\boxed{\bold{464~g}}

8 0
2 years ago
Neon has 2 isotopes. Neon-20 has a mass of 19:992 amu and
Alenkinab [10]

Answer:

20.0928.

Explanation:

The average atomic mass is (90 * 19.992 + 10* 21) / 100

= 20.0928.

6 0
2 years ago
The essential oil found in cloves, eugenol, can be isolated by steam distillation because it is insoluble in water and has a mea
Damm [24]
We know that:
Molar Mass H2O: 18 g/mol 
<span>Molar Mass of Eugenol: 164 g/mol </span>
<span>Boiling point of H2O: 100 degrees C </span>
<span>Boiling point of Eugenol: 254 degrees C </span>
<span>Density of water: 1.0 g/mL </span>
<span>Density of Eugenol: 1.05 g/mL </span>

<span>Using formula:
V= [mole fraction x molar mass] / density </span>

<span>mH20: 0.9947 * 18
          = 17.9046 / 1 g/mL
          = 17.9046 </span>
<span>morg: 0.0053 * 164  
        = 0.8692/ 1.05 g/mL
        = 0.8278 </span>

<span>V% = Vorg/(Vorg + VH2O) * 100 </span>
<span>(0.8278/18.7324) * 100 = 4.419% </span>

Yotal volume = 30 mL; therefore, 
<span>0.0442 = (volume eugenol/30) </span>

<span>(m eug/mH2O) = (peug*164/pH2O*18) </span>
<span>(m eug/30) = (4*164/760*18) </span>
<span>m eug = about 1.44g and </span>
<span>
volume = mass/density
            = 1.44/1.05
            = about 1.37 mL </span>
6 0
2 years ago
A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling
aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}        .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

We are given:

M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL

Putting values in above equation, we get:

0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g

Hence, the mass of glucose in final solution is 0.420 grams

3 0
3 years ago
1. A gas having the following composition is burnt under a boiler with 50% excess air.
jeka94

The composition of the stack gas are :

CH_4= 0.8713

C_3H_8 = 0.0202

CO = 0.107

<h3 /><h3>What is a mole fraction?</h3>

The ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the components.

Assuming 100 g of the stack gas. Calculate the mass of each species in this sample according to their percentages.

Mass of CH_4 : 70% of 100 g = 70 g

Mass of C_3H_8 : 15% of 100 g = 15 g

Mass of CO : 15% of 100 g = 15 g

Now calculate the number of moles of each species:

Number of moles of CH_4 : \frac{70 g}{16.04 g/mol} = 4.3 mole

Number of moles of C_3H_8: \frac{15 g}{144.1 g/mol} = 0.10 mole

Mass of CO : \frac{15 g}{28.01 g/mol} = 0.53 mole

Now to calculate the mole fraction of each we use the formula:

Mole fraction of CH_4: \frac{4.3}{4.935} = 0.8713

Mole fraction of C_3H_8 : \frac{0.10}{4.935} = 0.0202

Mole fraction of CO : \frac{0.53}{4.935} = 0.107

Hence, composition of the stack gas are:

CH_4 = 0.8713

C_3H_8 = 0.0202

CO = 0.107

Learn more about mole fraction here:

brainly.com/question/13135950

#SPJ1

8 0
2 years ago
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