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melamori03 [73]
1 year ago
7

Kajal is a compound or a mixture?​

Chemistry
1 answer:
luda_lava [24]1 year ago
8 0
It is a mixture…………….
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What ingredients are needed to turn the cake mix into cake batter​
svp [43]

Answer:

Ingredients

2 ⅓ cups all-purpose flour.

1 tablespoon baking powder.

¾ teaspoon salt.

1 ½ cups white sugar.

½ cup shortening.

2 eggs.

1 cup milk.

1 teaspoon vanilla extract.

Explanation:

4 0
2 years ago
If an object increases in speed, it must be as a result of
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DkskdjejisdjejkZjdjejasj
3 0
3 years ago
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Describe the relationship between the radius of a cation and that of the atom from which it is formed. a. cations are much small
koban [17]
<h3><u>Answer;</u></h3>

Cations are much smaller than their corresponding parent

<h3><u>Explanation;</u></h3>
  • Parent atom has more electrons and thus the effective nuclear charge on each electron is less.
  • When a cation is formed electron(s) is/are lost. Thus the effective nuclear charge or simply put, the attraction of the nucleus towards the electrons increases. Therefore, due to greater pull, the nucleus pulls the shells towards it, there by reducing the size, which makes cations smaller than their corresponding parent.
3 0
2 years ago
Which element has higher ionization energy? chlorine or carbon
PtichkaEL [24]

Answer:

chlorine has higher ionization than carbon

Explanation:

Chlorine is only one row below carbon, but it is three columns to the right in this case the IP of chlorine would be predicted to be greater than the IP of carbon.

3 0
3 years ago
A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

7 0
3 years ago
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