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2 years ago

7

Which is the following is a decomposition reaction

1 answer:

lora16 [44]2 years ago

5 0

The equation that is a decomposition reaction is

PbCO3(s) ---> PbO(s) + CO2(g)

That is option D.

<h3>Properties of a decomposition reaction</h3>

A chemical reaction is defined as the combination of substances called reactants to form a product. The various types of chemical reaction include:

synthesis reaction

combustion reaction

decomposition reaction

single or double replacement reaction

Decomposition reaction is that type of chemical reaction in which the reactant breaks down to form two or more products.

Therefore the heating of lead (II) carbonate is a decomposition reaction.

Learn more about decomposition reaction here:

brainly.com/question/15229375

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How much does 2.17 mole of CU weigh?

Sergio [31]

Answer:

137.9035 g

Explanation:

1 mole of copper in grams is 63.55 g

2.17 mole of Cu weighs 137.9035 g (63.55x2.17)

6 0

3 years ago

A block of aluminum occupies a volume of 15.0 cm and weighs 40.5 g. What is its density?

mafiozo [28]

Answer:

2.7 g/cm^3

Explanation:

D = m/V

D = 40.5 g / 15 cm^3

D = 2.7 g/cm^3

7 0

3 years ago

Perform the following conversions:

ololo11 [35]

Answer:

a) $20^0C$ and $293K$

b) $437K$ and $327.2^0F$

c) $-273^0C$ and $459.44^0F$

Explanation:

Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like $^0C$ , $^0F$ and $K$

These units of temperature are inter convertible.

$t^0C=(t+273)K$

$t^oC=\frac{5}{9}\times (t^oF-32)$

$K-273=\frac{5}{9}\times (^oF-32)$

a) 68°F (a pleasant spring day) to °C and K.

Converting this unit of temperature into $^0C$ and $K$ by using conversion factor:

$t^oC=\frac{5}{9}\times (68^oF-32)$

$t=20^0C$

$20^0C=(20+273)K=293K$

b) 164°C (the boiling point of methane, the main component of natural gas) to K and °F

Conversion from degree Celsius to Kelvins and Fahrenheit

$164C=\frac{5}{9}\times (t^oF-32)$

$t^0F=327.2$

$164°C=(164+273)K=437 K$

c) 0K (absolute zero, theoretically the coldest possible temperature) to °C and °F.

$K-273=\frac{5}{9}\times (t^oF-32)$

$0-273=\frac{5}{9}\times (t^oF-32)$

$t=-459.4^0F$

$t^K=(0-273)^0C=-273^0C$

8 0

3 years ago

Solutions can be composed of:

Anika [276]

A solution may exist in any phase so your answer is D. any of the above

hope this helps :)

7 0

3 years ago

God, cure my stupidity.

Cloud [144]

The answer is the third one

4 0

3 years ago

Read 2 more answers