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Fofino [41]
3 years ago
9

Read the following, and answer the questions below. Remember to read the question carefully.

Chemistry
1 answer:
andreev551 [17]3 years ago
8 0
<h3>Question #1:</h3>

False.

Melting is a type of chemical change.


<h3>Question #2:</h3>

True.

If the substance's atoms change in any way, it is a chemical change.


<h3>Question #3:</h3>

True.

If a substance is broken or torn, it's still the same substance. (Take tearing paper, for example)


<h3>Question #4:</h3>

False.

Take boiling water for example, when you boil water, it is still water, right?


<h3>Question #5:</h3>

False.

<em><u>Every</u></em> chemical or physical change in matter includes change in energy. <u><em>Not most.</em></u>


<h3>Question #6</h3>

True.

Hydrogen peroxide (H2O2), has turned into water (H2O) and oxygen (O). The substance has change chemically, therefore it is an chemical change.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I hoped this helps! If you have any questions, feel free to ask! ^^

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Solid lead IV oxide decomposes into solid lead and oxygen gas.
vredina [299]
The balanced equation for the decomposition of solid lead iv oxide is as follows: 2PbO2 = 2PbO + O2.
Lead IV oxide decompose to give lead ll oxide and oxygen. Lead iv oxide is thermally unstable and it usually decomposes into oxygen and lead ll oxide when heated. Lead ll oxide is more stable than lead lV oxide.
7 0
3 years ago
Given each pair, complete the sentences to determine which member of each has the stronger intermolecular dispersion forces.
evablogger [386]

Answer:

CH3CH2CH2Cl

CH3CH2CH2CH2CH2SH

Br2

Explanation:

Dispersion forces increases with increase in relative molecular mass. The specie having the greater relative molecular mass definitely has greater dispersion forces. A rough estimation of the relative molecular masses of the species stated in the answer will reveal this fact.

3 0
3 years ago
PLEASE HELP!!!! 20 MINS LEFT IM DYING PLS HELP
Zepler [3.9K]

Answer:

Kc = 0.20        

Explanation:

                  N₂O₄     ⇄    2NO₂

moles       5.3mol          2.3mol

Vol               5L                5L

Molarity    5.3/5M        2.3/5M

                = 1.06M     = 0.46M

Kc = [NO₂]²/[N₂O₄] =  (0.46)²/(1.06) = 0.1996 ≅ 0.20

6 0
3 years ago
Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, C_{Ag} = 78 wt%

concentration of Pd, C_P_d = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, A_A_g = 107.87 g/mol

atomic weight of iron, A_P_d = 106.4 g/mol

Step 1: determine the average density of the alloy

\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }

\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3

Step 2: determine the average atomic weight of the alloy

A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }

A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole

Step 3: determine unit cell volume

V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell

Step 4: determine the unit cell edge length

Vc = a³

a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0
3 years ago
11) Which group of Elements have Similar chemical Properties? *
Kryger [21]
Ca, Sr, Ba have similar chemical properties.
Explanation: They are in the same group (Alkaline Earth Metals)
6 0
3 years ago
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