What economic value do lemons bring to our nation

BRAINLIESTTT ASAP!!! PLEASE HELP ME :))

garri49 [273]

Answer:

Multiply 1.25 by 0.04 and divide the result obtained by 1,000

Explanation:

Given: [1 gram = 0.04 ounce, 1 liter = 1,000 milliliter]

1.25 x 0.04 = 0.05 oz

Therefore, 0.05 per 1,000 milliliter

0.05 ÷ 1,000 = 0.00005 oz

Therefore, the density of the gas is 0.00005 oz/mL

Hope this helps! :)

6 0

3 years ago

What is the product of silver nitrate + lead ii nitrate?

Kisachek [45]

Answer:

Explanation:

In theory, not much of anything. The vast majority of nitrates are water soluble. Aside, not sure what chemistry level you are at but you will probably be asked to know or memorize some solubility rules. This, for lack of a better phrase, Nitrate rule, is near spot on. With one exception—a rare one—all metal cationic nitrates are soluble in water. All of them. So, assuming you are talking about aqueous, water-based solutions of these salts and mixing them together, I expect nothing to occur. Both solutions, I believe are colorless in water and will thus remain so. If you had say a solution of Iron (III) nitrate and copper (II) nitrate, slightly different story. Both are colorful solutions and I would think you might see blending of colors but no reaction; no precipitate will form. You will probably learn about markers of a chemical reaction. One of these is a color change. Note, you should read this as a change of color from what you previously had. Going from red to blue or colorless to colored (or vice versa) is a strong indication of a reaction (e. g. evidence of bond-breaking and bond-formation). The mere mixing of colors does not constitute a chemical reaction.

3 0

3 years ago

Describe how the size of sediment particles affects their movement during deflation.

Oksi-84 [34.3K]

The larger the piece the longer it will take to break down. This is because it has more mass that needs to be broken down.

6 0

3 years ago

Write equations that show the processes that describe the first, second, and third ionization energies for a gaseous gadolinium

Inessa [10]

Answer:

Gd(g) → $Gd^{+}(g) + e^{-}$ , $Gd^{+}(g)$ → $Gd^{2+}(g) + e^{-}$ , $Gd^{2+}(g)$ → $Gd^{3+}(g) + e^{-}$

Explanation:

Ionization energy is the energy required to remove an electron from a gaseous atom or ion.

The first ionization energy is the energy required to remove the valence electron(outermost) from a neutral atom:

Gd(g)→ $Gd^{+}(g) + e^{-}$

The second ionization energy is the energy required to remove next/second electron from $x^{+}$ ion. The second ionization energy is always higher than the first: