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3 years ago

Why are large atoms more reactive than small atoms

Chemistry

1 answer:

n200080 [17]3 years ago

6 0

Answer:

large atoms have Valence electrons further from the nucleus and lose them more readily.

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In which pair do both compounds exhibit predominantly ionic bonding? A) KCl and CO2 B) SO2 and BaF2 C) F2 and N2O D) N2O3 and Rb

gtnhenbr [62]

Answer:

E) NaF and SrO

Explanation:

The ionic bonding occurs between atoms with a great difference in electronegativity. This usually happens between a metal and a non-metal.

<em>In which pair do both compounds exhibit predominantly ionic bonding? </em>

A) KCl and CO₂. NO. C and O are non-metals and present covalent bonding.

B) SO₂ and BaF₂. NO. S and O are non-metals and present covalent bonding.

C) F₂ and N₂O. NO. Both compounds contain non-metals and present covalent bonding.

D) N₂O₃ and Rb₂O. NO. N and O are non-metals and present covalent bonding.

E) NaF and SrO. YES. Na and Sr are metals while F and O are non-metals.

5 0

3 years ago

Write the technique used to seperated the following mixtures

SashulF [63]

Answer: chromatography

Explanation:

5 0

3 years ago

An ideal gas described by Ti=291K, Pi=1.50bar, and Vi=13.3L is heated at constant volume until P=15.0bar. It then undergoes a re

Salsk061 [2.6K]

Answer:

$W=-4601.4J$

Explanation:

Hello,

In this case, the steps are:

$291K,1.50bar, 13.3L \rightarrow 15.0bar 13.3L,T_2\rightarrow T_2, V_2, 1.50bar\rightarrow 291K,1.50bar, 13.3L$

In such a way, the work per mole (w) for that isothermal process turns out:

$w=RTln(\frac{P_1}{P_2} )=8.314\frac{J}{mol*K}*291K*\frac{1.50bar}{15.0bar} \\\\w=-5570.8\frac{J}{mol}$

In addition, if the moles are required, since it is an ideal gas:

$n=\frac{PV}{RT}=\frac{1.50bar*13.3L}{0.083\frac{bar*L}{mol*K}*291K} =0.826mol$

So the work is:

$W=-5570.8\frac{J}{mol} *0.826mol=-4601.4J$

Best regards.

8 0

3 years ago

Order the following elements from smallest to largest atomic radius:

Alexandra [31]

Answer:

F,O,N,C,B

Explanation:

SINCE THE ELECTRONEGATIVITY INCREASES ACROSS THE PERIOD THE ELECTRONS ARE TIGHTLY PACKED AROUND THE NEUCLEOUS MAKING THE RADIUS SMALL

7 0

3 years ago

Calculate the specific heat capacity of titanium if a 43.56 g sample absorbs 476 J as its temperature changes from 20.13 oC to 4

soldier1979 [14.2K]

The specific heat capacity of Titanium is found as 0.522 J / g°C.

<u>Explanation:</u>

We have to find the specific capacity using the formula as,

q = m × c × ΔT

Where,

q is the heat absorbed by the sample = 476 J

m is the mass of the sample = 43.56 g

c is the specific heat capacity of the sample = ?

ΔT is the temperature difference = 41.06° C - 20.13°C = 20.93° C

Now, we have to rewrite the equation to get the specific heat capacity as,

c = $$\frac{q}{m \times dT}$

= $$\frac{476 }{43.56\times 20.93}$

= 0.522 J / g°C

So the specific heat capacity of Titanium is found as 0.522 J / g°C.

6 0

3 years ago