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ZanzabumX [31]
3 years ago
12

Which of the following will increase the rate of disolving?

Chemistry
1 answer:
Fantom [35]3 years ago
3 0

Answer:

I have no idea bro

Explanation:

I feel you

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The term element refers to atoms that have the same number of _______.
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I guess it’s Electrons
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Take a look at this dandelion. The yellow flower on the right is _______ pollinated and the seeds on the left are transported by
bearhunter [10]

Answer:

gay bowsah

Explanation:

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A student wonders whether a piece of jewelry is made of pure silver. She determines that its mass is 3.25 g. Then she drops it i
nydimaria [60]

Answer:

The jewelry is 2896.54_Kg/m^3 less dense than pure silver

Explanation:

Density of jewellery = (mass of jewellery) ÷ (volume of jewellery)

=3.25g ÷ 0.428mL = 0.00325Kg÷0.000000428m^3 = 7583.46Kg/m^3

The density of silver is 10490_Kg/m^3 which is (10490 - 7583.46) 2896.54_Kg/m^3 more dense than the jewellery

The density of Silver [Ag]

The weight of Silver per cubic centimeter is 10.49 grams or the weight of silver per cubic meter is 10490 kilograms, that is the density of silver is 10490 kg/m³; at 20°C (68°F or 293.15K) at a pressure of one atmospheres.

7 0
3 years ago
Which equation shows how to calculate how many grams (g) of KOH would be needed to fully react with 4 mol Mg(OH)2? The balanced
bazaltina [42]
Mole ratio:

MgCl₂ + 2 KOH = Mg(OH)₂ + 2 KCl

2 moles KOH ---------------- 1 mole Mg(OH)₂
moles KOH ------------------- 4 moles Mg(OH₂)

moles KOH = 4 x 2 / 1

= 8 moles of  KOH

molar mass KOH = 56 g/mol

mass of KOH = n x mm

mass of  KOH = 8 x 56

= 448 g of KOH 

hope this helps!

3 0
3 years ago
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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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