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8_murik_8 [283]
3 years ago
11

Enter the chemical formula of a binary molecular compound of hydrogen and a Group 6A element that can reasonably be expected to

be more acidic in aqueous solution than H2S , e.g. have a larger Ka.
Chemistry
1 answer:
Ber [7]3 years ago
8 0

Answer:

H₂Se

Explanation:

A way of estimating the acidity of a weak acid is by analizing the<em> stability of the formed anion</em>. In this case, we should find a Group 6A element that in its anionic forms (HX⁻ and X⁻²) is more stable than HS⁻ and S⁻², thus it would be more acidic in aqueous solution.

The anionic forms of Se are more stable than the forms of S, similarly to how Br⁻ is more stable than Cl⁻.

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A single bond contains_____ <br> Shared electrons.<br><br> A.1<br><br> B.2
aleksley [76]

Answer:

<u>B. 2</u>

Explanation:

A single bond contains <u>2</u> shared electrons.

Given example is that of <u>ethane</u>

3 0
3 years ago
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What parts of a circuit is a conductor?
netineya [11]

Answer:

copper

Explanation:

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a calorimeter contained 75.0 g of water at 16.95 C. A 93.3-g sample of iron at 65.58 C was placed in it, giving a final temperat
Nostrana [21]

Answer:- \frac{382.69J}{0C} .

Solution:- Mass of Iron added to water is 93.3 g. Initial temperature of iron metal is 65.58 degree C and final temperature of the system is 19.68 degree C.

temperature change, \Delta T for iron metal = 65.58 - 19.68 = 45.9 degree C

specific heat for the metal is given as 0.444 J per g per degree C.

let's calculate the heat lost by iron metal using the equation:

q=mc\Delta T

where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:

q=93.3g(45.9^0C)(\frac{0.444J}{g.^0C})

q = 1901.42 J

Using same equation we will calculate the heat gained by water.

mass of water is 75.0 g.

\Delta T for water = 19.68 - 16.95 = 2.73 degree C

specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:

q=75.0g(\frac{4.184J}{g.^0C})(2.73^0C)

q = 856.674 J

Total heat lost by iron metal is the sum of heat gained by water and calorimeter.

So, heat gained by calorimeter = heat lost by iron metal - geat gained by water

heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J

Change in temperature for calrimeter is same as for water that is 2.73 degree C

For calorimeter, q=C.\Delta T

C=\frac{q}{\Delta T}

C=\frac{1044.746J}{2.73^0C}

C=\frac{382.69J}{0C}

So, the heat capacity of calorimeter is \frac{382.69J}{0C} .


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