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2 years ago

Question 2 of 19

Chemistry

1 answer:

Alborosie2 years ago

3 0

Answer:

Explanation:

The temperature increases

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Property of a substance

Scrat [10]

There are many properties to substances.

I'll list some examples below:

Mass
Volume
Density
Conductivity
Malleability
Boiling point
Melting point
Heat capacity

Hope this helps! :3

8 0

2 years ago

What type of reaction does the following equation represent?

Ostrovityanka [42]

Answer:

B. double-replacement RXN

Explanation:

more specifically, this is a precipitation rxn.

6 0

2 years ago

Read 2 more answers

Explain why BH3 is trigonal planar and NH3 is trigonal pyramidal.

Troyanec [42]

Answer:o It is important to realise that mixing will be small unless there are electrons in the 4a1 LUMO, this is why NH3 is pyramidal while BH3 is planar! ... This mixing is very strong and stabilises the 3a1 MO substantially and hence NH3 is trigonal pyramidal and not planar.

Explanation:

3 0

3 years ago

c3H8 + 5O2 = 3CO2 + 4H2O When 44.0 grams of propane (C3H8) under goes complete combustion, how many grams of water will be produ

Umnica [9.8K]

Answer: 72 grams of water will be produced.

Explanation:

To calculate the number of moles, we use the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

Mass of propane = 44 grams

Molar mass of propane = 44 grams

Putting values in above equation, we get:

$\text{moles of propane}=\frac{44g}{44g/mol}=1mole$

For the reaction of combustion reaction of propane, the equation follows:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

By Stoichiometry of the reaction,

1 mole of propane produces 4 moles of water.

So, 1 mole of propane will produce = $\frac{1}{1}\times 4=4moles$ of water.

Now, to calculate the amount of water, we use equation 1, we get:

Molar mass of water = 18 g/mol

$4mol=\frac{\text{Mass of water}}{18g/mol}$

Mass of water produced = 72 grams

Hence, 72 grams of water will be produced.

6 0

3 years ago

The first-order decay of radon has a half-life of 3.823 days. How many grams of radon remain after 7.22 days if the sample initi

Nesterboy [21]

substitute: t1/2=ln(2)k→k=ln(2)t1/2 Into the appropriate equation: [A]t=[A]0∗e−kt [A]t=[A]0∗e−ln(2)t1/2t [A]t=(250.0 g)∗e−ln(2)3.823 days(7.22 days)=67.52 g

6 0

3 years ago