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3 years ago

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When a chemical reaction occurs, matter is-----the amount of each atom is the reactants will----- be different form the amount i

n ----the. Are used to show how many of each molecule in present, and--------are used to show how many of each--------is present.

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Nimfa-mama [501]3 years ago

4 0

Answer:

plz explain more

Explanatin:

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A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

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

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

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

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

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

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

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3 years ago

Which sentence most accurately describes electrically charged objects?

ivolga24 [154]

Answer:

They are attracted to one other without coming into contact.

Explanation:

In conclusion, an electrically neutral object is an object that has a balance of protons and electrons. In contrast, a charged object has an imbalance of protons and electrons. The type of charge (positive or negative) is determined by whether the protons or the electrons are in excess

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Why is 0.15 M of hypochlorous acid acidic and 0.15 M of sodium hypochlorite basic?

yaroslaw [1]

As the name hypochlorous acid suggest that it is acidic, no matter what is the concentration of the acid, the acid solution is acidic as it provides hydronium or hydrogen ions to the aqueous solution.

HClO (aq) ↔ H⁺ (aq) + ClO⁻ (aq)

Hypochlorous acid or HClO is a weak acid, while NaOH is a strong base. The formation of NaOCl takes place when these two react. NaOCl is a salt of a strong base and a weak acid, and due to this, the solution of sodium hypochlorite is basic.

Sodium is the spectator ion, that is, it does not need to be represented at the time of demonstrating the net equation for the hydrolysis of sodium hypochlorite.

ClO⁻ (aq) + H2O (l) ↔ HOCl (aq) + OH⁻ (aq)

On observing the first equation, on the product side, there are hydrogen ions, which signifies that the solution is acidic, and if we observe at the second equation, on the product side hydroxide ion is witnessed that indicates that the solution is basic.

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4 years ago

In the process of performing a spectrophotometric determination of iron, an analyst prepares a calibration curve using a single-

Vesna [10]

Answer:

Yes.

Explanation:

Yes, the change in cuvette lead to a determinate error in the analysis if the different cuvette is used for the analysis because the amount of liquid sample that is used has different volume. If both cuvette are of the same type and has no difference in their structure and size then there is error occurs in the analysis but if both cuvette are different from one another then the error will occur in the analysis. because the amount of liquid that is used has different volume.

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3 years ago

What type of motion is illustrated in the diagram below?

Bezzdna [24]

You didn’t show the diagram

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