Answer:
The empirical formula of this substance is:
Explanation:
To find the empirical formula of this substance we need the molecular weight of the elements Carbon, Hydrogen and Oxygen, we can find this information in the periodic table:
- C: 12.01 g/mol
- H: 1.00 g/mol
- O: 15.99 g/mol
With the information in this exercise we can suppose in 100 g of the substance we have:
C: 48.64 g
H: 8.16 g
O: 43.2 g (100 g - 48.64g - 8.16g= 43.2 g)
Now, we need to divide these grams by the molecular weight:
We need to divide these results by the minor result, in this case O=2.70 mol
We need to find integer numbers to find the empirical formula, for this reason we multiply by 2:
This numbers are very close to integer numbers, so we can find the empirical formula as subscripts in the chemical formula:
I know this isn't the answer but here's an English translation
<span>The saturated vapor pressure of pure water at 20 ° C is 18 mmHg. If at the same temperature as much as 120 grams of x-substance dissolved in 990 grams of water, the vapor pressure of the solution becomes 17.37 mmHg, calculate the relative molecular mass of the substance x?</span>
The NMR is attached that is required to answer this question. We are told that we have a carboxylic acid and that there is a nitro group directly attached to an aromatic ring. We can begin by determining the substitution on the aromatic ring.
Looking at the NMR spectrum, we a peak that integrates to 1 H at 12 ppm which is characteristic of a carboxylic acid, which we already know is present. Next we have two equivalent doublets that both integrate to 2 H, giving us 4 hydrogens in total. These doublets are in the aromatic region and this type of coupling pattern is characteristic of a 1,4-substituted aromatic ring, so we know there is only one other group substituted on the ring. However, the molecular formula is C₉H₉NO₄, so there are still 2 carbons not accounted for, if we include our carboxlic acid. Therefore, the carboxylic acid must be attached to some alkyl group which is substituted onto the aromatic ring.
We have a doublet at 1.6 ppm that integrates to 3, which suggests this is a methyl group adjacent to a CH. We also have a quartet at 4.0 ppm with an integration of 1. This suggests it is a CH that is adjacent to 3 hydrogen, most likely the methyl group we just described.
Therefore, we have a CH attached to a CH3, so that CH requires two more bonds. The only pieces left to attach to it are the aromatic ring and the carboxylic acid functional group. This gives us the structure shown in the image provided.
Full Question:
A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?
How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?
2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)
Answer:
13.1 g K2CO3 required to neutralize spill
Explanation:
2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)
Number of moles = Volume * Molar Concentration
moles HBr= 0.42L x .45 M= 0.189 moles HBr
From the stoichiometry of the reaction;
1 mole of K2CO3 reacts with 2 moles of HBr
1 mole = 2 mole
x mole = 0.189
x = 0.189 / 2 = 0.0945 moles
Mass = Number of moles * Molar mass
Mass = 0.0945 * 138.21 = 13.1 g
Answer:
0.3229 M HBr(aq)
0.08436M H₂SO₄(aq)
Explanation:
<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>
<em />
Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).
NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)
When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.
<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>
<em />
Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).
2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)
When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.
