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3 years ago

In the presence of an emulsifying agent, a mixture of oil and water becomes a _____.

1 answer:

8 0

In the presence of an emulsifying agent, a mixture of oil and water becomes a colloidal dispersion.
Colloidal dispersion otherwise colloid is a system, in which discrete particles, droplets or bubbles of a dispersed phase (in this case oil), whose size at least in one dimension is in the range from 1 to 1000 nm are distributed in the other, usually continuous phase - dispersion medium (in this case water) differing from the dispersed phase in composition or state of aggregation.

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• Explain why each of the following pairs is not likely to form an ionic bond.

Olegator [25]

Answer:

A)Chlorine and Bromine:

They are both non metal hence they form a covalent bond due to covalent bonding.

B)Potassium and Helium:

Helium ion has a small cationic radius and distorted by the potassium ion due to polarization.

C)Sodium and Lithium:

Both are metals hence they form a metallic bond since they share electrons to the electron cloud.

4 0

3 years ago

PLEASE HELP!!! NEED TO PASS THIS TO THE FIRST SEMESTER!

Cloud [144]

Answer:

Reaction 1 is balanced but 2 is not balanced , the balance equation are :

1. $CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)$

2. $CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + H_{2}O(aq)$

Explanation:

Balanced Equations : These are the equation which follows the law of conservation of mass .

The total number of atoms present in reactant is equal to total number of atoms present in product.

1. $CH_{3}COOH(aq) + NaHCO_{3}(aq)\rightarrow CO_{2}(g) + H_{2}O(l) + CH_{3}COONa(aq)$

This is acid - base type reaction where

$CH_{3}COOH(aq)$ act as Acid

$NaHCO_{3}(aq)$ act as weak base

Reactant : $CH_{3}COOH(aq)$ , $NaHCO_{3}$

Number of atoms of :

C = 2 ( $CH_{3}COOH(aq)$ ) + 1 ( $NaHCO_{3}$ )

= 2 + 1

= 3

H = 4( $CH_{3}COOH(aq)$ ) + 1 ( $NaHCO_{3}$ )

= 4 + 1

O = 2( $CH_{3}COOH(aq)$ ) + 3 ( $NaHCO_{3}$ )

= 5

Na = 1 ( $NaHCO_{3}$ )

= 1

Product : $CO_{2}(g)$ , $H_{2}O(l)$ , $CH_{3}COONa(aq)$

Number of atoms :

C = 1( $CO_{2}(g)$ ) + 2( $CH_{3}COONa(aq)$ )

= 1 + 2

= 3

H = 2( $H_{2}O(l)$ ) + 3( $CH_{3}COONa(aq)$ )

= 2 + 3

= 5

O = 1( $H_{2}O(l)$ ) + 2( $CH_{3}COONa(aq)$ )

+2( $CO_{2}(g)$

= 1 + 2 + 2

= 5

Na = 1( $CH_{3}COONa(aq)$

= 1

Number of Na =1 , C = 3 , H= 5 and O =5 in both reactant and product , so it is a balanced reaction

2. $CaCl_{2}(aq) + 2NaHCO_{3}(aq)\rightarrow CaCO_{3}(aq) + H_{2}O(l) + 2NaCl(aq) + CO_{2}(g)$

This is double displacement reaction .

Check the balancing in both reactant and products should be :

Na = 2

H = 2

Ca = 1

C = 2

O = 6

Cl = 2

7 0

3 years ago

When methane is burned with oxygen, the products are carbon dioxide and water. if you produce 9 grams of water and 11 grams of c

Alina [70]

4 grams of methane is burned with oxygen,. Hope this helped

3 0

3 years ago

2. What group is the atom in? 2 or 2A 6 14 1

Juliette [100K]

Answer:

$\Large \boxed{\mathrm{Group \ 4}}$

Explanation:

The number of valence electrons tells us the group number of the neutral atom.

The atom has 4 valence electrons.

The atom is in group 4.

5 0

3 years ago

Read 2 more answers

What is the mass of an object with a density of 3.4 g/mL and a volume of 500.0 mL

umka2103 [35]

Formula:
$mass=density *volume$

Given:
Density=3.4
Volume=500.0

Plug them into the formula:
$mass=3.4 *500.0=1700$

Final answer: 1700g

5 0

3 years ago