In a scientific experiment, the factor that may change in response to the manipulated variable is called the _____

What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0

2 years ago

In a hospital laboratory, a 10.0 mL sample of gastric juice (predominantly HCl), obtained several hours after a meal, was titrat

LiRa [457]

Answer: 1.14

Explanation:

$HCl+NaOH\rightarrow NaCl+H_2O$

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?\\V_1=10.0mL\\n_2=1\\M_2=0.1M\\V_2=7.2mL$

Putting values in above equation, we get:

$1\times M_1\times 10.0=1\times 0.1\times 7.2\\\\M_1=0.072M$

To calculate pH of gastric juice:

molarity of $H^+$ = 0.072

$pH=-log[H^+]$

$pH=-log(0.072)=1.14$

Thus the pH of the gastric juice is 1.14

3 0

2 years ago

The enthalpy change for the explosion of ammonium nitrate with fuel oil is –7198 kJ for every 3 moles of NH4NO3. What is the ent

anastassius [24]

Answer:

−2399.33 kJ

Explanation:

If NH₄NO₃ reacts with fuel oil to give a ΔH of -7198 for every 3 moles of NH₄NO₃

What is the enthalpy change for 1.0 mole of NH₄NO₃ in this reaction

∴ For every 1 mole, we will have $\frac{1}{3}$ of the total enthaply of the 3 moles

so, to determine the 1 mole; we have:

$\frac{1}{3}*(-7198kJ)$

= −2399.33 kJ

∴ the enthalpy change for 1.0 mole of NH₄NO₃ in this reaction = −2399.33 kJ

7 0

2 years ago

Select the correct answer.

Phoenix [80]

Answer:

Molecular formula = C₆H₁₂O₆

Explanation:

Given data:

Mass of hydrogen = 31.7 g

Mass of carbon = 283.4 g

Mass of oxygen = 377.4 g

Molar mass of compound = 176.124 g/mol

Molecular formula = ?

Solution:

Number of gram atoms of H = 31.7 / 1.01 = 31.4

Number of gram atoms of O = 377.4 / 16 = 23.6

Number of gram atoms of C = 283.4 / 12 = 23.6

Atomic ratio:

C : H : O

23.6/23.6 : 31.4/23.6 : 23.6/23.6

1 : 1.33 : 1

C : H : O =3 (1 : 1.33 : 1 )

Empirical formula is C₃H₄O₃.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass = 3×12+4+3×16 = 88

n = 176.124 / 88

n = 12

Molecular formula = n (empirical formula)

Molecular formula = 2 (C₃H₄O₃)

Molecular formula = C₆H₁₂O₆

7 0

2 years ago

A gas effuses 1.55 times faster than propane (C3H8)at the

stepladder [879]

Answer: The mass of the gas is 18.3 g/mol.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{Rate_{X}}{Rate_{C_3H_8}}=1.55$

$\frac{Rate_{X}}{Rate_{C_3H_8}}=\sqrt{\frac{M_{C_3H_8}}{M_{X}}}$

$1.55=\sqrt{\frac{44}{M_{X}}$

Squaring both sides and solving for $M_{X}$

$M_{X}=18.3g/mol$

Hence, the molar mas of unknown gas is 18.3 g/mol.

5 0

2 years ago

Read 2 more answers

In a scientific experiment, the factor that may change in response to the manipulated variable is called the ______ variable.