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3 years ago

Determine the chemical formulas for the two compounds below. (Carbon, Hydrogen, and Oxygen).

Chemistry

1 answer:

tester [92]3 years ago

4 0

Remember that any intersection of lines is a C, and that the number of hydrogens attached are the necessary to complet the 4 bonds.

1) CH3 - CH (OH) - CH (CH3) -CH3

2) CH3 - O - CH(CH3)-CH2 - CH3

I have used the parenthesis to indicate that the radical inside is in other branch, bonded by a single line -

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The metal which does react vigorously with halogens to form halides?

pshichka [43]

Answer:

RUBIDIUM

Explanation:

5 0

3 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa

notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

$2NaOH+H_2SO_4-->Na_2SO_4+2H_2O$

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

$n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\$

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

$0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4$

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

$0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4$

Finally, the percent yield turns out into:

$Y=\frac{1.92g}{7.1g} *100$

$Y=27.0$ %

Best regards.

6 0

3 years ago

It was calculated that 4.3mL of 0.417 M HCl is required to titrate 11.9 mL of 0.151 M Mg(OH)2. Show evidence 2 HCl(aq) + Mg(OH)2

Lapatulllka [165]

Answer:

See explanation.

Explanation:

Hello,

In this case, for the described chemical reaction:

2 HCl(aq) + Mg(OH)2(aq) → MgCl2(aq) + 2 H2O(l)

We can notice there is a 2:1 molar ratio between the moles of hydrochloric acid and magnesium hydroxide, therefore, at the equivalence point:

$n_{HCl}=2*n_{Mg(OH)_2}$

And in terms of volumes and concentrations we verify:

$V_{HCl}M_{HCl}=2*V_{Mg(OH)_2}M_{Mg(OH)_2}$

So we use the given data to proof it:

$4.3mL*0.417M=2*11.9mL*0.151M\\1.793=3.594$

Therefore, we can conclude the data is wrong by means of the 2:1 mole ratio that for sure was not taken into account. This is also supported by the fact that normalities are actually the same, but the nomality of magnesium hydroxide is the half of the hydrochloric acid normality since the acid is monoprotic and the base has two hydroxyl ions.

Best regards.

4 0

3 years ago

If 10.0 moles of sulfur dioxide react with excess chlorine gas, what is the theoretical yield (in grams) of CI2O produced?

xz_007 [3.2K]

Answer:

869 g Cl₂O

Explanation:

To find the theoretical yield of Cl₂O, you need to (1) convert moles SO₂ to moles Cl₂O (via mole-to-mole ratio from reaction coefficients) and then (2) convert moles Cl₂O to grams Cl₂O (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs to reflect the sig figs of the given amount (10.0 moles).

1 SO₂ (g) + 2 Cl₂ (g) ----> 1 SOCl₂ (g) + 1 Cl₂O (g)

Molar Mass (Cl₂O): 2(35.453 g/mol) + 15.998 g/mol

Molar Mass (Cl₂O): 86.904 g/mol

10.0 moles SO₂ 1 mole Cl₂O 86.904 g
------------------------ x ---------------------- x ------------------ = 869 g Cl₂O
1 mole SO₂ 1 mole

6 0

2 years ago

Which graph shows a negative acceleration ( position and time )

allsm [11]

I'm pretty sure its graph 4

7 0

3 years ago