All categories

3 years ago

Which is a chemical property of iron

Chemistry

2 answers:

Gnesinka [82]3 years ago

7 0

Answer:

The chemical property of iron would be rusting in damp air

Explanation:

Whitepunk [10]3 years ago

5 0

a chemical property of iron is the ability to change it, hammer it, roll it, shape it etc

You might be interested in

What is the molar mass of HBR

alexdok [17]

Answer:

Molar mass of bromine is equal to $80.9119$

Explanation:

The molar mass of HBr is equal to the sum of atomic weight of Bromine.

Atomic Weight of hydrogen is equal to $1.00794$

Atomic Weight of Bromine is equal to $79.904$

Molar mass of Bromine

= Atomic Weight of hydrogen + Atomic Weight of Bromine

Molar mass of Bromine $= 79.904 + 1.00794 = 80.9119$

3 0

3 years ago

3. Explain what would happen to the digestion process if enzymes were not present. SC.6.L.14.5

Lesechka [4]

Without enzymes the nutrients in your food would go to waste which leads to a poor digestion and malnutrition. It’s because enzymes break down the nutrients which use it for energy

3 0

2 years ago

Porquê que o potássio é mais reativo que o cálcio? urgente!

saul85 [17]

Answer: O potássio possui apenas um elétron de valência. ... Considerando que o cálcio possui dois elétrons de valência, requer mais energia para a remoção de um elétron de valência. Devido a esse potássio, é mais reativo que o cálcio.

Explanation:

4 0

3 years ago

Name two things in the lab that are for safety

Blizzard [7]

Answer:

Safety goggles and chemical fume hoods

Explanation:

5 0

2 years ago

15. What volume of CCI, (d = 1.6 g/cc) contain

anastassius [24]

Answer:

$\boxed{\text{(3) 9.6 L}}$

Explanation:

1. Moles of CCl₄

$n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}$

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

$m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}$

4. Volume of CCl₄

$V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}$

4 0

2 years ago