Answer:
13.2 g of gold
Explanation:
We'll begin by converting 5.25 L to ft³.
This can be obtained as follow:
Recall:
1 L = 0.0353 ft³
Therefore,
5.25 L = 5.25 × 0.0353
5.25 L = 1.85×10¯¹ ft³
From the question given above,
2.45 g of gold is present in 5.25 L ( i.e 1.85×10¯¹ ft³) of soil.
Therefore, Xg of gold will be present in 1 ft³ of soil i.e
Xg of gold = 2.45/1.85×10¯¹
Xg of gold = 13.2 g
Therefore, 13.2 g of gold is present in 1 ft³ of the soil.
Answer:
continental? Bc it cover that one place within it's boundaries
No of atoms in 4.8 dm³ of neon gas is determined in the following way.
Explanation:
We have given
Neon Gas=4.8dm³
At STP
1 mole of an ideal gas occupies 22.4 L(dm³) of volume.
which means that number of molecules occupy 22.4 L of volume.
So
4.8 dm³x 1 mole/22.4 dm³ = 0.214 moles Neon
0.214 moles x 6.02x10²³ atoms/mole = 1.29x10²³ atoms of Neon
No of Atoms in 4.8dm3 neon gas =1.29x10²³ atoms of Neon
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Answer:
( About ) 5.7 grams
Explanation:
Take a look at the attachment below for a proper explanation;
Answer:
An emulsifying agent is typically characterized by having <u><em>d. one polar end and one nonpolar end.</em></u>
Explanation:
Emulsifiers are substances that have the ability to bind, for example, fats with those substances that have mostly water in their conformation. In other words, the emulsifier facilitates mixtures of two or more immiscible liquid substances.
This is because the molecules of an emulsifier are often lipophilic (attract oil) at one end and hydrophilic (attract water) at the other. In other words it consists of a polar (hydrophilic) head group and a non-polar (hydrophobic) tail.
<u><em>An emulsifying agent is typically characterized by having d. one polar end and one nonpolar end.</em></u>
