If 10 ml of orange juice has a mass of 5 g, what would the density of Orange

Chemistry

1 answer:

Sophie [7]3 years ago

5 0

Answer:

Sep by step to determine the mass of 25 mL of orange juice

Explanation:

Measure mass of container , then add 25 mL of OJ and subtract the difference

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What is a mole?

fredd [130]

Answer:

Explanation:

One mole is the amount of substance that contain the Avogadro number which is equal to 6.022×10^23 atom, molecules or ions.

7 0

3 years ago

Aqua regia, a mixture of HCl and HNO₃, has been used since alchemical times to dissolve many metals, including gold. Its orange

marissa [1.9K]

Aqua regia is an oxidative mixture that is highly corrosive and is composed of hydrochloric acid and nitric acid. The Ea (rev) for the reaction is 3 kJ.

<h3>What is activation energy?</h3>

The activation energy is the minimum required energy by the reactant to undergo changes to form the product. The activation energy of the reverse reaction is given by the difference in the production state and transition state.

It is given as,

Ea (rev) = Ea (fwd) − ΔHrxn

Given,

ΔH° = 83KJ

Ea (fwd) = 86 kJ/mol

Substituting the values above as:

Ea (rev) = 86 - 83

= 3 kJ

Therefore, the activation energy of the reverse reaction is 3 kJ.

Learn more about activation energy, here:

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4 0

1 year ago

If a patient has a medical condition that causes his cells to absorb fewer than normal molecules, this patient would likely feel

Svetach [21]

Answer:oxygen Explanation:The medical condition described here is anaemia. It is a blood cell disorder whereby the red blood cell doesn't function properly and hence doesn't carry enough oxygen to the tissues. This is usually caused when ones body is deficient of iron.The symptoms that may occur to such patients are weakness, fatigue, headache and pale skin.Based on the explanation, the answer is oxygen

Explanation:

5 0

2 years ago

A solution may contain Ag+, Pb2+, and/or Hg22+. A white precipitate forms when 6 M HCl is added. The precipitate is partially so

omeli [17]

Answer:

All three are present

Explanation:

Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: $AgCl (s), PbCl_2 (s), Hg_2Cl_2 (s)$ .

Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
Secondly, addition of liquid ammonia would form a precipitate with silver: $AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq)$ ; Silver hydroxide at higher temperatures decomposes into black silver oxide: $2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g)$ .
Thirdly, we also know we have $Hg_2^{2+}$ in the mixture, since addition of potassium chromate produces a yellow precipitate: $Hg_2^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow Hg_2CrO_4 (s)$ . The latter precipitate is yellow.