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Zielflug [23.3K]
1 year ago
8

List the bonding pairs H and I; S and O; K and Br; Si and Cl, H and F; Se and S; C and H in order of increasing covalent charact

er. 1. (S,O) < (Se,S) < (C,H) = (H,I) = (H,F) < (Si,Cl) < (K,Br) 2. (C,H) = (H,I) = (Se,S) < (K,Br) < (S,O) = (Si,Cl) < (H,F) 3. (K,Br) < (Si,Cl) < (H,F) = (H,I) = (C,H) < (Se,S) < (S,O) 4. (K,Br) < (H,F) < (Si,Cl) < (S,O) < (H,I) = (C,H) < (Se,S) 5. (Se,S) < (C,H) < (H,I) < (S,O) < (Si,Cl) = (H,F) < (K,Br) 6. (H,F) < (Si,Cl) = (S,O) < (K,Br) < (Se,S) = (H,I) = (C,H) 7. None of these
Chemistry
1 answer:
stellarik [79]1 year ago
5 0

Answer:

1. (S,O) < (Se,S) < (C,H) = (H,I) = (H,F) < (Si,Cl) < (K,Br)

Explanation:

The covalent character always increases down the group, this is because ionic character decreases down the group and also electronegativity.

In the same way, Covalent character always decreases across a period because electronegativity increases across a period.

The higher the electronegativity values between the two atoms, the more ionic it will be.

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Copper atoms are heavier than magnesium atoms. So, although each atom of magnesium can produce one atom of copper, the masses wo
Ilya [14]

Answer:

m_{Cu}=0.99gCu

Explanation:

Hello!

In this case, considering the given chemical reaction and the mass of the magnesium strip, following the indications of the atomic weight ratio (2.61 g Cu/1 g Mg), and keeping in mind the 1:1 mole ratio one could compute the produced mass of copper as shown below:

m_{Cu}=0.38gMg*\frac{2.61gCu}{1gMg} \\\\m_{Cu}=0.99gCu

Best regards!

8 0
1 year ago
What is [h3o ] in a solution of 0.25 m ch3co2h and 0.030 m nach3co2?
Brilliant_brown [7]
Hello!

To determine [H₃O⁺], we need to apply the Henderson-Hasselback equation, since this is a case of an acid and its conjugate base:

pH=pKa+log( \frac{[A^{-}] }{[HA]} )

pH=4,76+log( \frac{0,030M}{0,25M} ) \\  \\ pH=3,84

Now, we use the definition of pH and clear [H₃O⁺] from there:

pH=-log[H_3O^{+}]

[H_3O^{+}] = 10^{-pH} =10^{-3,84}=0,00014 M

So, the [H₃O⁺] concentration is 0,00014 M

Have a nice day!
4 0
2 years ago
If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________. If the initial concen
Nikolay [14]

This is an incomplete question, here is a complete question.

If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________.

The reaction

2NOBr(g)\rightarrow 2NO(g)+Br_2(g)

It is a second-order reaction with a rate constant of 0.80 M⁻¹s⁻¹ at 11 °C.

Answer : The concentration of after 9.0 seconds is, 0.00734 M

Explanation :

The integrated rate law equation for second order reaction follows:

k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)

where,

k = rate constant = 0.80 M⁻¹s⁻¹

t = time taken  = 142 second

[A] = concentration of substance after time 't' = ?

[A]_o = Initial concentration = 0.0440 M

Putting values in above equation, we get:

0.80M^{-1}s^{-1}=\frac{1}{142s}\left (\frac{1}{[A]}-\frac{1}{(0.0440M)}\right)

[A]=0.00734M

Hence, the concentration of after 9.0 seconds is, 0.00734 M

5 0
1 year ago
The gaseous product of a reaction is collected in a 25.0-l container at 27
nataly862011 [7]
To find the number of moles of gas we can use the ideal gas law equation, we dont need to use the mass of gas given as we only have to find the number of moles 
PV = nRT 
P - pressure - 300.0 kPa 
V - volume - 25.0 x 10⁻³ m³
n - number of moles 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 27 °C + 273 = 300 K
substituting these values in the equation 
300.0 kPa x 25.0 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 300 K 
n = 3.01 mol 
number of mols of gas - 3.01 mol
4 0
1 year ago
Explain why the spectrum produced by a 1-gram sample of element Z would have the same spectral lines at the same wavelengths as
TEA [102]
The reason why the spectrum produced by a 1 gram sample of element z would have the same spectral lines at the same wavelengths as the spectrum produced by a 2gram is because  Mass does not determine the spectral lines

hope this helps
6 0
2 years ago
Read 2 more answers
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