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zubka84 [21]
3 years ago
15

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 , e.g. have a larger .
Chemistry
1 answer:
rosijanka [135]3 years ago
5 0

The question is incomplete, the complete question is;

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.

Answer:

H2Se

Explanation:

The acidity and extent of acid dissociation (shown by Ka value) depends on the nature of the bond between hydrogen and the group 6A element.

If we compare the H-S bond and the H-Se bond, we will discover that the H-Se bond is longer, weaker and breaks more easily than the H-S bond. As a result of this, H2Se is more acidic than H2S.

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Does a reaction occur when aqueous solutions of silver(I) acetate and manganese(II) iodide are combined?
san4es73 [151]

Answer:

When aqueous solutions of silver(I) acetate and manganese(II) iodide are combined, an insoluble precipitate of silver(I) iodide is formed

Explanation:

  • It is an example of precipitation reaction.
  • When aqueous solutions of silver(I) acetate and manganese(II) iodide are combined, an insoluble precipitate of silver(I) iodide is formed.
  • Precipitation of silver(I) iodide is confirmed by it's yellow color.
  • Hence a reaction is observed.
  • Molecular reaction: 2Ag(OAc)(aq.)+MnI_{2}(aq.)\rightarrow 2AgI(s)+Mn(OAc)_{2}(aq.)
7 0
3 years ago
A gas mixture consists of 4 kg of O2, 5 kg of N2, and 7 kg of CO2. Determine (a) the mass fraction of each component, (b) the mo
kondor19780726 [428]

Answer:

See explanation.

Explanation:

Hello

(a) In this case, one uses the following formulas, which allow to compute the mass fraction of each component:\% O_2=\frac{m_{O_2}}{m_{O_2}+m_{N_2}+m_{CO_2}}*100\%=\frac{4kg}{4kg+5kg+7kg}*100\%=25\%O_2\\\% N_2=\frac{m_{N_2}}{m_{O_2}+m_{N_2}+m_{CO_2}}*100\%=\frac{5kg}{4kg+5kg+7kg}*100\%=31.25\%N_2\\\% CO_2=\frac{m_{CO_2}}{m_{O_2}+m_{N_2}+m_{CO_2}}*100\%=\frac{7kg}{4kg+5kg+7kg}*100\%=43.75\%CO_2

(b) For the mole fractions, it is necessary to find all the components' moles by using their molar mass as shown below:

n_{O_2}=4kgO_2*\frac{1kmolO_2}{32kgO_2} =0.125kmolO_2\\n_{N_2}=5kgN_2*\frac{1kmolN_2}{28kgN_2} =0.179kmolN_2\\n_{CO_2}=7kgCO_2*\frac{1kmolCO_2}{44kgCO_2} =0.159kmolCO_2

Now, the mole fractions:

x_{O_2}=\frac{n_{O_2}}{n_{O_2}+n_{N_2}+n_{CO_2}}*100\%=\frac{0.125kmolO_2}{0.125kmolO_2+0.179kmolN_2+0.159kmolCO_2}*100\%=27\%O_2\\x_{N_2}=\frac{n_{N_2}}{n_{O_2}+n_{N_2}+n_{CO_2}}*100\%=\frac{0.179kmolN_2}{0.125kmolO_2+0.179kmolN_2+0.159kmolCO_2}*100\%=38.7\%N_2\\ x_{CO_2}=\frac{n_{CO_2}}{n_{O_2}+n_{N_2}+n_{CO_2}}*100\%=\frac{0.159kmolCO_2}{0.125kmolO_2+0.179kmolN_2+0.159kmolCO_2}*100\%=34.3\%CO_2

(c) Finally the average molar mass is computed considering the molar fractions and each component's molar mass:

M_{average}=0.27*32g/mol+0.387*28g/mol+0.343*44g/mol=34.57g/mol

And the gas constant:

Rg=0.082\frac{atm*L}{mol*K}*\frac{1mol}{34.57g}=0.00237\frac{atm*L}{g*K}

Best regards.

8 0
4 years ago
14. Name each of the following acids:<br> a. HF<br> d. H2SO4<br> b. HBr<br> e. H2PO4<br> C. HNO,
k0ka [10]

Answer: a. Hydrofluoric acid

d.sulfuric acid

e. Dihydrogen phospataion

b.hydrobromic acid

c. Nitric acid

Explanation:

6 0
4 years ago
Tellurium has eight isotopes: Te-120 (0.09%), Te-122 (2.46%), Te-123 (0.87%), Te-124 (4.61%), Te-125 (6.99%), Te-126 (18.71%), T
Elena-2011 [213]

The average atomic mass of tellurium, calculated from its eight isotopes (Te-120 (0.09%), Te-122 (2.46%), Te-123 (0.87%), Te-124 (4.61%), Te-125 (6.99%), Te-126 (18.71%), Te-128 (31.79%), and Te-130 (34.48%)) is 127.723 amu.

The average atomic mass of Te can be calculated as follows:

A = m_{Te-120}\%_{Te-120} + m_{Te-122}\%_{Te-122} + m_{Te-123}\%_{Te-123} + m_{Te-124}\%_{Te-124} + m_{Te-125}\%_{Te-125} + m_{Te-126}\%_{Te-126} + m_{Te-128}\%_{Te-128} + m_{Te-130}\%_{Te-130}

Where:

m: is the mass

%: is the abundance percent

Knowing all the masses and abundance values, we have:

A = 120*0.09\% + 122*2.46\% + 123*0.87\% + 124*4.61\% + 125*6.99\% + 126*18.71\% + 128*31.79\% + 130*34.48\%

To find the <u>average atomic mass</u> we need to change all the <u>percent values</u> to <u>decimal ones</u>

A = 120*9 \cdot 10^{-4} + 122*2.46 \cdot 10^{-2} + 123*8.7\cdot 10^{-3} + 124*4.61 \cdot 10^{-2} + 125*6.99\cdot 10^{-2} + 126*0.1871 + 128*0.3179 + 130*0.3448 = 127.723

Therefore, the average atomic mass of tellurium is 127.723 amu.

You can find more about average atomic mass here brainly.com/question/11096711?referrer=searchResults

I hope it helps you!

4 0
3 years ago
Determine the empirical formula of the following compound if a sample contains 0.104 molK, 0.052 molC, and 0.156 molO.
Nadya [2.5K]

The empirical formula is K₂CO₃.  

The empirical formula is the <em>simplest whole-number ratio of atoms</em> in a compound.  

The ratio of atoms is the same as the ratio of moles, so our job is to calculate the <em>molar ratio of K:C:O</em>.  

I like to summarize the calculations in a table.  

<u>Element</u> <u>Moles</u>  <u>Ratio</u>¹ <u>Integers</u>²  

     K       0.104   2.00         2

     C       0.052  1.00          1

     O      0.156   3.00         3

¹ To get the molar ratio, you divide each number of moles by the smallest number.  

² Round off the number in the ratio to integers to integers (2, 1, and 3).

The empirical formula is K₂CO₃.

6 0
3 years ago
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