A resistor has a voltage Across it of 3kv and a current through it of 3m A .

Abdid is an astronomer who has been observing objects that orbit the Sun in the asteroid belt. He finds a previously undiscovere

Elza [17]

Since it is round and it is the asteroid belt it is most likely that he found a dwarf planet

6 0

3 years ago

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To a 0.0001 m solution of mg(no3)2, naoh was added to a final concentration of 0.001m did a precipitate form?

Natalija [7]

I looked on a solubility chart to answer this question, and hydroxides are generally insoluble (with some exceptions of course). However, it says to consider $Mg(OH)_{2}$ as an insoluble substance, though it may be moderately soluble.

The answer that you are most likely looking for is: Yes, a precipitate does form - this is due to the double placement reaction:

$Mg(NO_{3})_{2}_{(aq)}$ + $2NaOH_{(aq)}$ → $Mg(OH)_{2} {(s)}$ + $2NaNO_{3}_{(aq)}$

8 0

3 years ago

Un tanque de acetileno para una antorcha de soldadura de oxiacetileno proporciona 9340 L de gas acetileno, C2H2, a 0°C y 1 atm 2

Helen [10]

Answer:

3.33 tanques de O₂

Explanation:

Basados en la reacción:

2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(g)

2 moles de acetileno reaccionan con 5 moles de oxígeno produciendo 4 moles de dióxido de carbono y 2 moles de agua

La ley de Avogadro dice que el volumen de un gas bajo temperatura y presión constantes es proporcional a las moles de este gas. Así, como 2 moles de acetileno reaccionan con 5 moles de oxígeno, los litros de O₂ necesarios para quemar 9340L de acetileno son:

9340 L C₂H₂ × (5 moles O₂ / 2 moles C₂H₂) = 23350L de O₂

Si un tanque contiene 7x10³ L de O₂ serán necesarios:

23350L O₂ ₓ (1 tanque / 7x10³L) = 3.33 tanques de O₂

6 0

3 years ago

Select the compound that is most likely to increase the solubility of ZnSe when added to water.NaCnMgBr2NaClKClO4

vlada-n [284]

Answer:

NaCl.

Explanation:

In the solution, ZnSe ionizes to $Zn^2^+$ and $Se^2^-$ . Following reaction represents the ionization of ZnSe in solution -

$ZnSe$ ⇄ $Zn^2^+ + Se^2^-$

As we want to increase the solubility of ZnSe, we must decrease the concentration of dissociated ions so that the reaction continues to forward direction.

If we add NaCl to this solution, then we have $Na^+$ and $Cl^-$ in the solution which will be formed by the ionization of NaCl.

Now, $Zn^2^+$ in the solution will react with two $Cl^-$ ions to form $ZnCl_2$ as follows -

$Zn^2^++2Cl^-$ ⇄ $ZnCl_2$

Due to this reaction the concentration of $Zn^2^+$ will decrease in the solution and more ZnSe can be soluble in the solution.

6 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.

3 0

3 years ago