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svp [43]
3 years ago
8

What would you use if you needed to determine the density of an object?

Chemistry
1 answer:
11Alexandr11 [23.1K]3 years ago
4 0
You would need a scale of some kind
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Using a case scenario describe how you will use the scientific method to conduct an investigation within your community
kenny6666 [7]
Depends what kind of case we taking about
5 0
3 years ago
Write the precipitation reaction for cobalt(II) hydroxide in aqueous solution: Be sure to specify the state of each reactant and
Len [333]

Answer:

The equation for the precipitation reaction of cobalt (ii) hydroxide is given below:

CoSO₄ (aq) + NaOH (aq) ----> Co(OH)₂ (s) + Na₂SO₄ (aq)

Explanation:

Cobalt (ii) hydroxide is an inorganic compound consisting of cobalt (ii) ions, Co²+ and hydroxide ions, OH-. It is insoluble in water and the pure form known as the beta form is a pink-coloured solid. The impure form which incorporates other anions in its molecular structure is blue in colour and is ustable.

Cobalt (ii) hydroxide is formed as precipitate when an alkaline metallic hydroxide such as sodium hydroxide is mixed with an aqueous cobalt (ii) salt such as cobalt (ii) sulfate. The equation for the precipitation reaction of cobalt (ii) hydroxide is given below:

CoSO₄ (aq) + NaOH (aq) ----> Co(OH)₂ (s) + Na₂SO₄ (aq)

Being a basic hydroxide, cobalt (ii) hydroxide neutralizes acids to form cobalt (ii) salts and water. For example: Co(OH)₂ (s) + H₂SO₄ (aq) ---> CoSO₄ (aq) + H₂O

Thus, cobalt (ii) hydroxide is soluble in acids.

Cobalt(II) hydroxide is used mostly as a drying agent for paints, varnishes, and inks. It is also useful in the preparation of other cobalt compounds.

7 0
3 years ago
Aluminum oxide breaks down into it elemental components when electric current is applied.
Sav [38]

Answer:

The primary use of aluminum oxide is the manufacture of aluminum metal. When an electric current is passed through molten (melted) aluminum oxide, the compound breaks down to form aluminum metal and oxygen gas.

Explanation:

8 0
2 years ago
P
BARSIC [14]

5.426 x 10^{22} molecules are there in a deep breath of air whose volume is 2. 35 L at body temperature 36 C and a pressure of 740 torr.

Explanation:

Data given:

volume V = 2.35 L

Temperature T = 36 Degrees or 273.15 + 36 = 309.15 K

pressure P = 740 torr OR 0.973 Atm

R = 0.08205 L atm / mole K

n (number of molecules of gas)= ?

The formula used will be

PV = nRT

Putting the value of given variables in above equation:

n = \frac{PV}{RT}

 = \frac{0.973 x 2.35}{0.08205 x 309.15}

 = 0.0901 moles

According to Avagadro' s law that gases at equal volume, temperature and pressure contains same molecules.

number of molecules:

number of moles x avagadro number

0.0901 x 6.02 x 10^{23}

= 0.54 x 10^{23}

= 5.426 x 10^{22} molecules.

6 0
3 years ago
An experiment requires 40.0g of ethylene glycol, a liquid whose density is 1.114 g/mL. Rather than weigh the sample on a balance
uranmaximum [27]
Density= mass/volume
volume=mass/density
volume= 40.0g/1.114g per mL
volume= 35.90664273 mL
volume = 35.9 mL
6 0
3 years ago
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