Which is not a characteristic of minerals

1 answer:

6 0

I think d is wrong because Everything else is correct for sure. also, minerals vary in the position they hold and have different chemical compounds.

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Easy Questions! Chemistry - Please help! I give thanks

Evgen [1.6K]

This is not chemistry but

see it is a triangular prim on it's side
V=BH
are of base times height
the base is a triangle
height is 18.5

base=1/2bh
b=8.6
h=8.4
base=1/2(8.6)(8.4)
base=36.12

V=bh
V=36.12*18.5
V=668.22

round
V=668.2 ft^3

8 0

3 years ago

Read 2 more answers

Two species of frog mate in the same pond. One breeds in early summer and one in late summer. This is an example of what kind of

MatroZZZ [7]

Answer:

Pre-zygotic, temporal separation

Explanation:

Reproductive isolation mechanism is of two types:

Prezygotic mechanism
Postzygotic mechanism

Prezygotic mechanism isolation occurs before fertilization and helpful in preventing formation of fertile offspring.

In frog external fertilization occurs. In the external fertilization, eggs and sperms are released in water and fertilization occur outside the water.

Prezygotic isolating mechanisms may include behavioral isolation, temporal isolation, mechanical isolation, gametic isolation and habitat isolation.

Temporal separation in reproduction is the sexual activity in the same geographical range but in different periods.

Therefore, the given reproductive isolation is pre-zygotic, temporal separation.

4 0

3 years ago

How many moles of sodium hydroxide would react with 1 Mole of sulphuric acid?

Paul [167]

Answer:

Two moles.

Explanation:

Sulphuric (sulfuric) acid $\rm H_2SO_4$ is a diprotic acid. When one mole of $\rm H_2SO_4$ molecules dissolve in water, two moles of $\rm H^{+}$ ions would be produced.

$\rm H_2SO_4 \to 2\, H^{+} + {SO_4}^{2-}$ .

On the other hand, sodium hydroxide $\rm NaOH$ is a monoprotic base. When one mole of $\rm NaOH$ formula units dissolve in water, only one mole of hydroxide ions $\rm OH^{-}$ would be produced.

$\rm NaOH \to Na^{+} + OH^{-}$ .

Note that $\rm H^{+}$ and $\rm OH^{-}$ react at a one-to-one ratio:

$\rm H^{+} + OH^{-} \to H_2O$ .

As a result, it would take $2\; \rm mol$ of $\rm OH^{-}$ to react with the $\rm 2\; mol$ of $\rm H^{+}$ that was released when $1\; \rm mol$ of $\rm H_2SO_4$ is dissolved in water. Since one mole of $\rm NaOH$ formula units could produce only one mole of $\rm OH^{-}$ , it would take $\rm 2\; mol$ of $\rm NaOH$ formula units to produce that $2\; \rm mol$ of $\rm OH^{-}$ for reacting with $1\; \rm mol$ of $\rm H_2SO_4$ .