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Leokris [45]
2 years ago
6

does a positive charge make the atomic radius smaller? and does a negative charge make the atomic radius bigger? yes or no.

Chemistry
1 answer:
Goryan [66]2 years ago
4 0
The answer to the question is no it doesn’t
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Formation of water: 2H2 + 1 O2 --> 2H20
VLD [36.1K]

Answer:

n=15 moles H2O

1:2

x:15

x=7.5 moles of O2

3 0
2 years ago
Can someone pls help me
Alex777 [14]

Answer:

the picture is blurd

Explanation:

7 0
2 years ago
When a teacher dilutes 50 mL of 2.0 M NaOH to 0.50 M, what volume of NaOH results?
JulijaS [17]

Answer:

= 200 mL

Explanation:

Using the dilution formula;

M1V1 = M2V2 ;

Where, M1 is the concentration before dilution, V2 is the volume before dilution, while M2 is the concentration after dilution and V2 is the volume after dilution.

M1 = 2.0 M

V1 = 50 mL

M2 = 0.50 M

V2 = ?

V2 = M1V1/M2

     = ( 2.0 × 50 )/ 0.5

     = 200 mL

Therefore, the volume after dilution will be, 200 mL

6 0
3 years ago
In this reaction, a gaseous system with a volume of 3.00 L has a rate of formation of NOCl of 0.0120 M s-1 . The volume of the s
oee [108]

Answer:

x = 0.324 M s⁻¹

Explanation:

Equation for the reaction can be represented as:

 2 NO(g)   + Cl₂ (g)      ⇄     2NOCl (g)

Rate = K [NO]² [Cl₂]

Concentration = \frac{numbers of mole (n)}{volume (v)}

from the question; their number of moles are constant since the species are quite alike.

As such; if Concentration varies inversely proportional to the volume;

we have: Concentration ∝ \frac{1}{v}

Concentration = \frac{1}{v}

Similarly; the Rate can now be expressed as:

Rate = K [NO]² [Cl₂]

Rate = (\frac{1}{v}) ^2 (\frac{1}{v} )

Rate = \frac{1}{v^3}

We were also told that the in the reaction, the gaseous system has an initial volume of 3.00 L and rate of formation of 0.0120 Ms⁻¹

So we can have:

0.0120 = \frac{1}{3^3}

0.0120 = \frac{1}{27}   -----Equation (1)

Now; the new rate of formation when the  volume of the system decreased to 1.00 L can now be calculated as:

x = (\frac{1}{1})^3

x = 1             ------- Equation (2)

Dividing equation (2) with equation (1); we have:

\frac{0.0210}{x} = \frac{\frac{1}{27} }{1}

\frac{0.0210}{x} = \frac{1}{27}

x = 0.0120 × 27

x = 0.324 M s⁻¹

∴  the new rate of formation of NOCl = 0.324 M s⁻¹

8 0
3 years ago
Radical's of
aivan3 [116]

Answer:

1..CO3

2..OH

3..CO

4..Cl

5..PO3

6..SO4

8..NO3

9..NO2

1O..I2

8 0
3 years ago
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