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ludmilkaskok [199]
3 years ago
12

PLZ HELP ME..ITS MY TEST

Chemistry
2 answers:
Sati [7]3 years ago
8 0

Answer: Oxidizing Agent

Explanation: When Cr is reduced (gains electrons), it is oxidizing the other element (removing its electrons) so it is the oxidizing agent.

artcher [175]3 years ago
4 0
Oxidizing agent would be the best answer choice i would choose
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Combustion _____.
KIM [24]
In combustion energy is released so it is an exothermic process !!

the answer is 2nd one !!
3 0
3 years ago
Read 2 more answers
Please solve this for points please dont just write random stuff<br><br> 40 points! and brainliest
Zepler [3.9K]

there is not context random people cant solve this because there is no way to know what the chart means and also don't fake 40 points. I'm sorry but if you can add context I might be able to help :/

5 0
3 years ago
Calculate the mass of NaF that must be added to 300ml if a 0.25m HF solution to form a buffer solution with a ph of 3.5​
Kamila [148]

Answer:

6.574 g NaF into 300ml (0.25M HF) => Bfr with pH ~3.5

Explanation:

For buffer solution to have a pH-value of 3.5 the hydronium ion concentration [H⁺] must be 3.16 x 10⁻⁴M ( => [H⁺] = 10^-pH = 10⁻³°⁵ =3.16 x 10⁻⁴M).

Addition of NaF to 300ml of 0.25M HF gives a buffer solution. To determine mass of NaF needed use common ion analysis for HF/NaF and calculate molarity of NaF, then moles in 300ml the x formula wt => mass needed for 3.5 pH.

HF ⇄ H⁺ + F⁻; Ka = 6.6 x 10⁻⁴

Ka = [H⁺][F⁻]/[HF] = 6.6 x 10⁻⁴ = (3.16 x 10⁻⁴)[F⁻]/0.25 => [F⁻] = (6.6 x 10⁻⁴)(0.25)/(3.16x10⁻⁴) = 5.218M in F⁻ needed ( = NaF needed).

For the 300ml buffer solution, moles of NaF needed = Molarity x Volume(L)

= (5.218M)(0.300L) = 0.157 mole NaF needed x 42 g/mole = 6.574 g NaF needed.

Check using the Henderson - Hasselbalch Equation...

pH = pKa + log ([Base]/[Acid]); pKa (HF) = 3.18

Molarity of NaF = (6.572g/42g/mole)/(0.300 L soln) = 0.572M in NaF = 0.572M in F⁻.

pH = 3.18 + log ([0.572]/[0.25]) ≅ 3.5.

One can also back calculate through the Henderson -Hasselbalch Equation to determine base concentration, moles NaF then grams NaF.

4 0
3 years ago
At the end of an experiment, the product is a mixture of the starting material, which is benzoic acid and the product, which is
Pani-rosa [81]

Answer:

b. Extraction with a solution of sodium hydrogen carbonate, separating the layers, followed by drying and evaporating the organic layer.

Step-by-step explanation:

The ether solution contains your product, benzaldehyde, and some starting material, benzoic acid, the purification steps are:

  1. Extract with a solution of NaHCO₃. The ether layer contains benzaldehyde, and the aqueous layer contains sodium benzoate.
  2. Separate the layers. Keep the ether layer.
  3. Dry the ether solution.
  4. Distill the ether (boiling point 35 °C) and purify the benzaldehyde (178 °C) by steam distillation.

a. is wrong. Extraction with HCl will not remove much of the benzoic acid.

c. is wrong. If you evaporate the organic layer, you will have a mixture of benzaldehyde and benzoic acid.

d. is wrong. If you work with the aqueous layer, you will end up with benzoic acid,

7 0
3 years ago
Determine the mass of 5.2 x 10 power of 21 molecules of propanol C3H7OH(l), on grams.
Alja [10]

n(\text{C}_3\text{H}_7\text{OH}) = N(\text{C}_3\text{H}_7\text{OH}) / N_A\\ \phantom{n(\text{C}_3\text{H}_7\text{OH})} = 5.2 \times 10^{21} / (6.02 \times 10^{23})\\ \phantom{n(\text{C}_3\text{H}_7\text{OH})} = 8.6 \times 10^{-3} \; \text{mol}

where N_A = 6.02 \times 10^{23}\; \text{mol}^{-1} the Avogadro's constant that relates the number of particles to their number, in the unit moles \text{mol}.

The molar mass of propanol- mass per mole propanol- can be directly deduced from its molecular formula with reference to a modern periodic table.

M(\text{C}_3\text{H}_7\text{OH}) = \underbrace{3 \times 12.01}_{\text{carbon}} + \underbrace{8 \times 1.008}_{\text{hydrogen}} + \underbrace{1\times 16.00}_{\text{oxygen}} = 60.09 \; \text{g} \cdot \text{mol}^{-1}

8.6 \times 10^{-3} \; \text{mol} of propanol molecules would thus have a mass of 8.6 \times 10^{-3} \; \text{mol} \times 60.09 \; \text{g} \cdot \text{mol}^{-1} = 0.52 \; \text{g}

5 0
4 years ago
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