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labwork [276]
3 years ago
11

Copy and balance these chemical equstions​

Chemistry
2 answers:
Rzqust [24]3 years ago
8 0
Is there supposed to be a question?
lyudmila [28]3 years ago
4 0

where are equations dear????

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HELP PLEASE! :( 
BabaBlast [244]

Answer:

Part 1: W = 116 Y = 163

Part 2: Since 232 is the mailing point of 2 kg then you would divide 232 by 2 to get the melting point for 1 kg, the same with Y.

6 0
4 years ago
Read 2 more answers
Why do surfers put wax on their boards​
poizon [28]

Answer:

to keep the surfer from slipping off the board

Explanation:

Surfboard wax (also known as surfwax) is a formulation of natural and/or synthetic wax for application to the deck of a surfboard, bodyboard, or skimboard, to keep the surfer from slipping off the board when paddling out or riding a wave. It is also used to increase grip on the paddle of a surf kayak or dragon boat.

5 0
3 years ago
Read 2 more answers
Why should you pay attention to the physical states of reactants and products when writing elquilibrium constant expressions?
babunello [35]
Answer is: because pure liquids (<span>shown in </span>chemical reactions<span> by appending (</span>l)<span> to the </span>chemical formula) and solids (<span>shown in </span>chemical equations by appending (s)<span> to the </span>chemical formula) not go in to he equilibrium constant expression, only gas state (shown in chemical reactions by appending (g) to the chemical formula) reactants and products go in to he equilibrium constant expression.
For example, equilibrium constant expression Kp for reaction:
A(s) + 2B(s) ⇄ 4C(g) + D(g).<span>
will be: Kp = [C]</span>⁴<span>·[D].
But for reaction </span>A(g) + 2B(g) ⇄ 4C(g) + D(g), will be:<span>
Kp = [C]</span>⁴<span>·[D] / [A]·[B]².</span>
3 0
4 years ago
How many atoms are there in 14.3 g of xenon?
olasank [31]

Answer:

6.56 × 10²² atoms Xe

Explanation:

Step 1: Find conversions

Molar Mass Xe - 131.29 g/mol

Avagadro's Number: 6.022 × 10²³

Step 2: Use Dimensional Analysis

14.3 \hspace{3} g \hspace{3} Xe(\frac{1 \hspace{3} mol \hspace{3} Xe}{131.29 \hspace{3} g \hspace{3} Xe} )(\frac{6.022(10)^{23} \hspace{3} atoms \hspace{3} Xe}{1 \hspace{3} mol \hspace{3} Xe} ) = 6.55911 × 10²² atoms Xe

Step 3: Simplify

We have 3 sig figs.

6.55911 × 10²² atoms Xe ≈ 6.56 × 10²² atoms Xe

8 0
4 years ago
How much of the original material would be left if it undergone 1 half lofe?
Zigmanuir [339]
Plz specify your answer. thank u
8 0
3 years ago
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