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zhenek [66]
3 years ago
8

Name two chemicals that can be used to prepare a standard solution in the Laboratory​

Chemistry
1 answer:
pantera1 [17]3 years ago
6 0

Answer:

Two chemicals that can be used to prepare standard solutions in lab are Hydrochloric acid & Sulphuric acid.

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Can anyone help me with this moon TGA?
kherson [118]

Answer:

i wont le me down load it

Explanation:

8 0
3 years ago
A rock weighing 26.0g placed graduated cylinder displacing the volume from 13.12ml to 25.3ml what is the density of the rock in
tia_tia [17]

density = mass /volume

1)mass of rock = 26.0 gram

2)volume of rock = volume of water it has displaced= 25.3-13.12 = 12.18

put the value of mass and volume in first equation and get the density value

3 0
3 years ago
Convert mass to moles for both reactants. (Round to 2 significant figures.)
shepuryov [24]
Mass = mr x moles
Mr of CuCl2 = ( 63.5) + ( 35.5 x 2) = 134.5
2.5 = 134.5 x moles
2.5 / 134.5 = moles
Moles = 0.019 (2DP)

0.25g of Al
Mr of Al = 27
0.25 = 27 x moles
0.25/ 27 = 0.0093 moles (2sf)

Hope this helps :)
6 0
3 years ago
Read 2 more answers
Explain why mixing of red paint with white paint does not constitute a chemical reaction even though the product has a different
MArishka [77]

Answer:

Explanation:

In a chemical change through chemical reactions, a new kind of matter forms. When a paint is mixed together, a physical change has occurred and not a chemical change. The different color one perceives is just a mere overlap between the reflected white and red color which produces another hue.

Not all color changes infers a chemical reaction. A paint is a complex mixture on its own.

5 0
3 years ago
Determine the specific heat ofmaterial if a 12g sample absorbed 48j as it was heated from 20-40
devlian [24]

Answer:

c =0.2 J/g.°C

Explanation:

Given data:

Specific heat of material = ?

Mass of sample = 12 g

Heat absorbed = 48 J

Initial temperature = 20°C

Final temperature = 40°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT =  40°C -20°C

ΔT =  20°C

48 J = 12 g×c×20°C

48 J =240 g.°C×c

c = 48 J/240 g.°C

c =0.2 J/g.°C

6 0
3 years ago
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