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1 year ago

In the formula X2O5, the symbol X could represent an element in Group

Chemistry

1 answer:

topjm [15]1 year ago

6 0

Answer: (3) 15

Explanation: We criss-cross down the oxidation numbers to get the subscripts for the correct formulas. That means the X has an oxidation number of 5. The element with the + oxidation number is always written first so it is +5. Of the groups names, only group 15 has +5 as an oxidation number.

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Which sentences describe alpha particles? Check all that apply 1. Alpha particles consist of two protons and two neutrons 2. Alp

Masja [62]

thanks for the answers ッ. (btw they’re on the bottom of the question if anyone doesn’t see it.

4 0

1 year ago

Read 2 more answers

Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?

kramer

Answer:

Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?

a. Al (s)

b. H2O (l)

c. HCN (g)

d. CH3COOH (l)

e. C2H6 (g)

Explanation:

Entropy is the measure of the degree of disorderness.

In solids, the entropy is very less compared to liquids and gases.

The entropy order is:

solids<liquids<gases

Among the given substances, water in liquid form has a strong intermolecular H-bond.

So, it has also less entropy.

Next acetic acid.

Between the gases, HCN, and ethane, ethane has more entropy due to very weak intermolecular interactions.

HCN has slight H-bonding in IT.

Hence, the entropy order is:

Al(s) < CH3COOH (l) <H2O(l) < HCN(g) < C2H6(g)

7 0

1 year ago

A sample of gas initially occupies 3.35 L at a pressure of 0.950 atm at 13.0oC. What will the volume (in L) be if the temperatur

Marat540 [252]

Answer: V= 3.13 L

Explanation: solution attached:

Use combine gas law equation:

P1 V1 / T1 = P2 V2/ T2

Derive to find V2

V2 = P1 V1 T2 / T1 P2

Convert temperatures in K

T1= 13.0°C + 273 = 286 K

T2= 22.5°C + 273 = 295.5 K

Substitute the values.

4 0

2 years ago

a compound contain 8.57g of carbon and 1.43g of hydrogen. The relative formula mass is 70 calculate the empirical formula of the

yanalaym [24]

Answer:

Empirical formula: CH2

Molecular formula: C5H10

5 0

1 year ago

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

8 0

1 year ago