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ikadub [295]
2 years ago
13

Balance the following chemical reaction. ____NaOH+____H2SO4 → ___Na2SO4+____H2O

Chemistry
1 answer:
kobusy [5.1K]2 years ago
7 0
The balanced equation would be 2NaOH + H2SO4 → Na2SO4 + 2H2O
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An unknown metal displaces Ali from solution but does not displace sodium using the activity series determine the unknown metal
MAXImum [283]

Answer:

The reactivity of metal is determined by the reactivity series. ... The metal which easily displaced aluminium will lie above in the series but that same element cannot displace sodium, so it will lie below in the series. Hence, from the series, we conclude that the unknown metal could be calcium or magnesium.

Explanation:

Hope this helps! :)

8 0
2 years ago
What element produces violet light of approximately 412 nm?
Natali [406]
a. Hydrogen

Hydrogen has a line at 410 nm.
Mercury has a line at 405 nm.
Sodium and neon have no lines near 412 nm.
8 0
2 years ago
Read 2 more answers
Which of the following is a galvanic cell?
sladkih [1.3K]

C. Aluminum (Al) oxidized, zinc (Zn) reduced

<h3>Further explanation</h3>

Given

Metals that undergo oxidation and reduction

Required

A galvanic cell

Solution

The condition for voltaic cells is that they can react spontaneously, indicated by a positive cell potential.

\large {\boxed {\bold {E ^ ocell = E ^ ocatode -E ^ oanode}}}

or:  

E ° cell = E ° reduction-E ° oxidation  

For the reaction to occur spontaneously (so that it E cell is positive), the  E° anode must be less than the E°cathode

If we look at the voltaic series:

<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>

The standard potential value(E°) from left to right in the voltaic series will be greater, so that the metal undergoing an oxidation reaction (acting as an anode) must be located to the left of the reduced metal (as a cathode)

<em />

From the available answer choices, oxidized Al (anode) and reduced Zn (cathode) are voltaic/galvanic cells.

7 0
3 years ago
Read 2 more answers
The reducing agent in this catalytic hydrogenation reaction was molecular hydrogen (H2), which was produced in situ (in the reac
creativ13 [48]

Answer:

a)  After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) 4NaBH_{4} + 2HCl + 7H_{2}O ⇒ 16H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7} c) No H_{2} was not the limiting reactant.

Explanation:

Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.

a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.

b)  4NaBH_{4} + 2HCl + 7H_{2}O ⇒ 16H_{2(g)}+ 2NaCl + Na_{2}B_{4}O_{7}

c) H_{2} was not the limiting reactant based on the mol to mol ratio of H_{2} and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of H_{2} would also be produced.

4 0
3 years ago
Two atoms of calcium and one molecule of oxygen gas react and yield two molecules of calcium oxide.
Vitek1552 [10]
2Ca + O2 ——————-> 2CaO2
3 0
3 years ago
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