Complete and balance the single-replacement reactions Mg + NiSO₄→

Can someone help me with my stoichiometry homework, it is due tonight.

AleksandrR [38]

Answer:

1. 1 mole of NaCl.

2. 1 mole of H₂O.

Explanation:

The balanced equation for the reaction is given below:

HCl + NaOH —> NaCl + H₂O

From the balanced equation above,

1 mole of HCl reacted with 1 mole of NaOH to produce 1 mole of NaCl and 1 mole of H₂O.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above, we can see that 1 mole of HCl reacted completely with 1 mole of NaOH. Therefore, HCl is the limiting reactant and NaOH is the excess reactant since we have 3.6 moles of NaOH.

1. Determination of the amount of NaCl produced from the reaction.

NOTE: The limiting reactant is used to determine the maximum amount produced since all of it is consumed in the reaction.

The limiting reactant is HCl and the amount of NaCl produced can be obtained as follow:

From the balanced equation above, we can see that 1 mole of HCl reacted to 1 mole of NaCl.

Thus, 1 mole of NaCl was produced.

2. Determination of the amount of H₂O produced from the reaction.

From the balanced equation above, it is evident that 1 mole of HCl reacted to produce 1 mole of H₂O.

Thus, 1 mole of H₂O was produced.

6 0

3 years ago

What is the molar mass of MnP

igomit [66]

The production of manganese peroxidase (MnP) by Irpex lacteus, purified to electrophoretic homogeneity by acetone precipitation, HiPrep Q and HiPrep Sephacryl S-200 chromatography, was shown to correlate with the decolorization of textile industry wastewater. The MnP was purified 11.0-fold, with an overall yield of 24.3%. The molecular mass of the native enzyme, as determined by gel filtration chromatography, was about 53 kDa. The enzyme was shown to have a molecular mass of 53.2 and 38.3 kDa on SDS-PAGE and MALDI-TOF mass spectrometry, respectively, and an isoelectric point of about 3.7. The enzyme was optimally active at pH 6.0 and between 30 and 40 degrees C. The enzyme efficiently catalyzed the decolorization of various artificial dyes and oxidized Mn (II) to Mn (III) in the presence of H(2)O(2). The absorption spectrum of the enzyme exhibited maxima at 407, 500, and 640 nm. The amino acid sequence of the three tryptic peptides was analyzed by ESI Q-TOF MS/MS spectrometry, and showed low similarity to those of the extracellular peroxidases of other white-rot basidiomycetes.

5 0

4 years ago

Using complete subshell notation (1s22s22p6, and so forth), predict the electron configuration of each of the following atoms: a

masha68 [24]

Answer: a) $C:6:1s^22s^22p^2$

b) $P:15:P:1s^22s^22p^63s^23p^3$

c) $Sb:1s^22s^22p^63s^23p^64s^23d^{10}4p^64d^{10}5s^25p^3$

Explanation:

Electronic configuration represents the total number of electrons that a neutral element contains. We add all the superscripts to know the number of electrons in an atom.

The electrons are filled according to Afbau's rule in order of increasing energies and thus the electronic configuration:

a) $C:6:1s^22s^22p^2$

b) $P:15:P:1s^22s^22p^63s^23p^3$

c) $Sb:1s^22s^22p^63s^23p^64s^23d^{10}4p^64d^{10}5s^25p^3$

4 0

4 years ago

Partner Namets) For each of the following reactions carried out: Write the balanced chemical equation, the full ionic equation,

torisob [31]

Answer : The full balanced ionic equation will be,

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)$

Reactants are lead nitrate and potassium iodide.

Products are lead iodide and potassium nitrate.

The spectator ions are, $K^+,NO_3^-$

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When potassium iodide react with lead nitrate then it gives potassium nitrate and lead iodide as a product.

The full balanced ionic equation will be,

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)$

The ionic equation in separated aqueous solution will be,

$2K^+(aq)+2I^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow PbI_2(s)+2K^+(aq)+2NO_3^{-}(aq)$

In this equation, $K^+\text{ and }NO_3^-$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Pb^{2+}(aq)+2I^{-}(aq)\rightarrow PbI_2(s)$

4 0

4 years ago

What happens when the elements in group 2 react with water?

attashe74 [19]

Answer:

The Group 2 metals become more reactive towards the water as you go down the Group.

Explanation:

These all react with cold water with increasing vigour to give the metal hydroxide and hydrogen. ... You get less precipitate as you go down the Group because more of the hydroxide dissolves in the water. Summary of the trend in reactivity.

Please mark me brainliest! hope this helped!

God bless!

8 0

3 years ago