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Ede4ka [16]
2 years ago
6

Given that the electronic configuration of an element X is 1s2 2s2 2p6 3s2 3p4,it can be deduced that X:

Chemistry
1 answer:
Maksim231197 [3]2 years ago
6 0

Answer:

Option A

Explanation:

The element X belongs to

=> Period 3 (Because it's valence shell is 3)

=> Group VI (Because electrons in the valence shell , 2+4 = 6)

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Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress
ElenaW [278]

Answer:

(a) W_{isoentropic}=8.125\frac{kJ}{mol}

(b) W_{polytropic}=7.579\frac{kJ}{mol}

(c) W_{isothermal}=5.743\frac{kJ}{mol}

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant k should be computed for air as an ideal gas by:

\frac{R}{Cp_{air}}=1-\frac{1}{k}  \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\

0.2856=1-\frac{1}{k}\\\\k=1.4

Next, we compute the final temperature:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K

Thus, the work is computed by:

W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}

(b) In this case, since n is given, we compute the final temperature as well:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K

And the isentropic work:

W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}

Regards.

8 0
2 years ago
A spherical balloon initially contains 25m3 of helium gas at 20o C and 150 kPa. A valve is now opened and the helium is allowed
Tpy6a [65]

Find the attachment for complete solution

6 0
2 years ago
What mass of sodium chloride contains 4.59 x 10 24 formula units?
shtirl [24]
The term formula units means molecules.

Then, what you are looking for is the mass in 4.59*10^24  molecules.

The procedure involves to convert the 4.59 * 10^24 molecules into moles and use the molar mass of the sodium chloride.

1) Number of moles = 4.59 * 10^24 molecules / (6.02 * 10^23 molecules/mol) = 7.62 mol

2) Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol

3) mass of NaCl = molar mass *  number of moles = 58.44 g/mol * 7.62 mol = 445.31 g of NaCl

Answer: 445.31 g of NaCl.

 
7 0
2 years ago
Joan’s initial nickel (II) chloride sample was green and weighed 4.3872 g. After the dehydration reaction and removal of excess
ANEK [815]

Answer:

a) yes, it was an hydrate

b) the number of waters of hydration, x = 6

Explanation:

a) yes it was an hydrate because the mass decreased after the process of dehydration which means removal of water thus some water molecules were present in the sample.

b) NiCl2. xH2O

mass if dehydrated NiCl2 = 2.3921 grams

mass of water in the hydrated sample = mass of hydrated - mass of dehydrated = 4.3872 - 2.3921 = 1.9951 g which represent the mass of water that was present in the hydrated sample.

NiCl2.xH2O

mole of dehydrated NiCl2 = m/Mm = 2.3921/129.5994 = 0.01846 mole

mole of water = m/Mm = 1.9951/18.02 = 0.11072 mole

Divide both by the smallest number of mole (which is for NiCl2) to find the coefficient of each

for NiCl2 = 0.01846/0.01846 = 1

for H2O = 0.11072/0.01846 = 5.9976 = 6

thus the hydrated sample was NiCl2. 6H2O

4 0
3 years ago
What is biochemistry​
lana66690 [7]

Answer:

Branch of science concerned with the chemical and physicochemical processes and substances that occur within living organisms.

7 0
2 years ago
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