All categories

3 years ago

As a trend or pattern,the atomic radius go from left to right across the period but top to bottom down a group

1 answer:

5 0

As a trend or pattern, the atomic radius decreases as you go left to right across the period, but increases as you go down a particular group.

You might be interested in

How many moles are in 8.32 x 1024 molecules of CO2.

Andrews [41]

Answer:

<h2>13.82 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{8.32 \times {10}^{24} }{6.02 \times {10}^{23} } \\ = 13.820598...$

We have the final answer as

<h3>13.82 moles</h3>

Hope this helps you

6 0

3 years ago

A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of

Bad White [126]

Answer:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

$\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

The number of moles of the ideal gas is:

$n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}$

$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$

$n = 5.541\,mol$

The final temperature is:

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}$

$T_{2} = 360.955\,K$

The final pressure is:

$P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}$

$P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)$

$P_{2} = 138569.171\,Pa\,(1.386\,bar)$

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

$\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}$

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}$

$T_{2} = 347.348\,K$

The final volume is:

$V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}$

$V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})$

$V_{2} = 0.14\,m^{3}$

4 0

4 years ago

How many atoms of oxygen are in 4.00 moles of N2O5

SVETLANKA909090 [29]

Answer:

there are 20 oxygen atoms in 4.00 moles of Dinitrogen pentoxide

Explanation:

there are 2 atoms in an oxygen molecule , so each oxygen molecules has at least 2. Dinitrogen pentoxide is N2O5, which has 7 atoms, 2 nitrogen and 5 oxygen. 1 molecule of N2O5 has 5 oxygen atoms, so 4 of then would be 20

8 0

3 years ago

Which statement is true about a pseudoscientific idea? It is biased in its results. It can be tested and observed. It can be rep

vesna_86 [32]

<span>It is biased in its results.

</span>

6 0

4 years ago

Read 2 more answers

eleanor purchased $2568 worth of stock and paid her broker a 0.5% fee. She sold the stock when the stock price increased to 3928

Mrrafil [7]

Answer: $1338.16

Explanation: Total cost of stock= $2568

Total cost of stock including the brokerage $=2568+\frac{0.5}{100}\times {2568}=2580.84$ $

Selling price of stock = $3928

Selling price of stock including trading fee=($3928-$7)=$3919

Net Proceeds = Net selling price of stock - Cost Price of stock

Net Proceeds = ($3919 - $2580.84) = $1338.16

5 0

3 years ago

As a trend or pattern,the atomic radius go from ______ left to right across the period but ______ top to bottom down a group

As a trend or pattern,the atomic radius go from left to right across the period but top to bottom down a group