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alexdok [17]
3 years ago
6

why mixture of stone pebbles And water is concerned considered suspension you are provided with a mixture of sand salt and camph

or how would you separate these particles explain in detail​
Chemistry
1 answer:
baherus [9]3 years ago
6 0

Answer:

camphor sublimates salt is soluble in water while sand does not sublime and does not dissolve in water you first heat the mixture in a beaker covered with a watch glass camphor will then accumulate on the watch glass then you dissolve the remaining mixture of sand and salt salt will dissolve forming a salt solution then you filter using a filter paper and a beaker the residue on the filter paper is sand while the filtrate is salt solution you then heat the salt solution so that it can evaporate leaving salt particles thus you will have obtained salt sand and camphor

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If 50. 75 g of a gas occupies 10. 0 l at stp, 129. 3 g of the gas will occupy ________ l at stp.
NeX [460]

22.4L

of any gas contains 1 mol of that gas.

50.75g/10L*22.4L/1 mol= 113.68g/mol- this is the mole weight of your gas

1 mol/113.68g*129.3g=1.137403 mol

Set up a ratio

1.137403mol/x L=1 mol/22.4 L

X=25.477827L, or with sig figs, x=25.5L

8 0
2 years ago
Balance the following Chemical Equation:<br> NaBr +CaCl2-&gt; NaCl+ CaBr2
frozen [14]
First write all of the compounds/atoms in either side then fill in existing values and balance


Na- 1
Br- 1
Ca- 1
Cl- 2

Na- 1
Cl- 1
Ca-1
Br-2

Balance to get

2NaBr+CaCl2=2NaCl+CaBr2
7 0
3 years ago
(6 points) Calculate the maximum number of moles and grams of H2S that can form when 158 g of aluminum sulfide reacts with 131 g
Schach [20]

Answer:107.1 g, 124.1 g

Explanation:

The equation of the reaction is;

Al2S3(s) + 6H20(l) ----> 2Al(OH)3(s) + 3H2S(g)

Hence;

For Al2S3

Number of moles= reacting mass/molar mass

Number of moles = 158g/150gmol-1 =1.05 moles

If 1 mole of Al2S3 yields 3 moles of H2S

1.05 moles of Al2S will yield

1.05 × 3/1 = 3.15 moles

Mass of H2S = 3.15moles × 34 gmol-1 = 107.1 g

For water

Number of moles of water = 131g/18gmol-1= 7.3 moles

6 moles of water yields 3 moles of H2S

7.3 moles of water will yield 7.3 × 3/6 = 3.65 moles of H2S

3.65 moles × 34 gmol-1 =124.1 g

5 0
3 years ago
Someone please help me and answer my question asap. i need to turn my assignment in soon... please.
Ksenya-84 [330]

Answer:

Explanation:

a. the salt produced would be Mg3N2(magnesium nitride)

b. magnesium loses 2 electron to form Mg2+ ion and nitrogen gains 3 electron to form n3-

when several of these ions come together 3 Mg2+ ion combine with 2 n3- ion to form Mg3N2 thus Mg getting six electron from nitrogen to form a ionic bond.

c. the reaction is not balanced Mg + N2 = Mg3n2

to make it balanced the reaction should be 3 Mg + N2 = Mg3N2.

the reaction was not balanced before because the number of Mg on both side of the reaction was not equal.

d. magnesium nitrate has formula Mg(NO3)2 is formed when Mg combines with nitrogen and oxygen Mg + N2 + o2

8 0
3 years ago
8. Select the lattice energy for rubidium chloride from the following data (in kJ/mol]
yKpoI14uk [10]

Answer:

Option C

Explanation:

The chemical reactions which are involved while solving this problem is there in the file attached and each chemical reaction is represented by a certain equation number

Lattice energy for rubidium chloride ( RbCl) is represented by the equation 6

Equation 1 represents the change in enthalpy for formation of RbCl

Equation 2 represents the sublimation reaction of rubidium

Equation 3 represents the ionization enthalpy of rubidium

Equation 4 represents the enthalpy of atomization of chlorine which means it describes the bond enthalpy of Cl2 molecule

Equation 5 represents the electron affinity of chlorine

To find the lattice energy for RbCl we have to use all the equations from 1 to 5 so that at last we get the equation 6

We have to perform operations such as

Equation 1 - equation 2 - equation 3 - equation 4 - equation 5

By performing these operations the intermediate compounds gets cancelled and at last we get equation 6

So Equation 1 ≡  ΔH_{f} = -431 kJ/mol

Equation 2 ≡ Rb(s) ---> Rb(g) = 85.8  kJ/mol

Equation 3 ≡ IE1(Rb) = 397.5  kJ/mol

Equation 4 ≡ BE(Cl2) = 226  kJ/mol

Equation 5 ≡ Electron Affinity Cl = -332  kJ/mol

Value corresponding to the equation 6 will be the value of lattice energy of RbCl and the value is -695·3 kJ/mol

∴ Lattice energy for rubidium chloride is approximately -695 kJ/mol

4 0
3 years ago
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