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n200080 [17]
3 years ago
15

Because they are not bonded it is much ________________ to separate the molecules

Chemistry
1 answer:
dalvyx [7]3 years ago
7 0
Hydrolysis I think
I’m not sure if its correct
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What is the mass of a sample containing 1.94 moles of Ca? ______g
Andrej [43]

Answer:

77.8 grams of Ca

Explanation:

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What are 2 chemical changes of water?​
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Hydrogen peroxide in water and adding kool- aid powder to water so the liquid turns red
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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.

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3 years ago
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Elements consist of tiny particles called
Nutka1998 [239]
The answer is atoms good luck
4 0
3 years ago
Q2.  0.254 g of KHP (204 g/mol) titrated against 20.0 mL of unknown NaOH (40.0 g/mol) solution to get the end point of phenolpht
Vera_Pavlovna [14]

Answer:

<u>Mass concentration (g/L) </u><u><em>= 2.49g/L.</em></u>

Explanation:

No. of moles = \frac{mass}{molar mass}

= \frac{0.254}{204} = 0.001245 moles

Concentration of KHP (C1) in litres = n/v

= \frac{0.001245}{0.02} = 0.062 mol/L

We know that:

C_{1} V_{1} = C_{2} V_{2}

where c1v1 and c2v2 are the products of concentration and volumes of KHP and NaOH respectively.

Since mole ratio is 1 : 1.

1 mole of NaOH - 40g

0.001245 mole of NaOH = 40 × 0.001245 = 0.0498g

⇒0.0498g of NaOH was used during the titration

<u><em>∴Mass concentration (g/L) = 0.0498g ÷ 0.02L</em></u>

<u><em>= 2.49g/L.</em></u>

3 0
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