All categories

2 years ago

Help pls.!! It's due tonight

Chemistry

1 answer:

marshall27 [118]2 years ago

8 0

Answer:

1. It will most likely die

2. The ostrilope with level 10 armor will most likely reproduce more offspring because it is adapted to the environment.

You might be interested in

Both prokaryotes and eukaryotes have small, free-floating organelles. The cell makes these organelles from nucleic acids and ami

serg [7]

Answer:

C. ribosomes

Explanation:

Both prokaryotic and eukaryotic cells contain certain structures called ORGANELLES. They possess some in common and others are not found in one or the other. According to this question, a small, free-floating organelle made from nucleic acid and amino acid is found in both eukaryotes and prokaryotes. This organelle is RIBOSOMES.

Ribosomes are organelles responsible for the synthesis of protein in both eukaryotes and prokaryotes. They can be found free-floating or attached to endoplasmic reticulum. Ribosomes are predominantly made of RNA (nucleic acid) and proteins i.e. Ribosomal RNA (rRNA) and proteins is their structural constituent. Hence, the organelle in this question is RIBOSOME.

5 0

2 years ago

Will the final amount of biodiesel change if you actually have more or less amount of catalyst

DerKrebs [107]

Answer

I don know sorry

Explanation:

4 0

2 years ago

If the first isotope has a mass of 259.98 amu and a percent abundance of 35%, what is the percent abundance of the second isotop

brilliants [131]

If it is assumed that there are only two isotopes then the percent abundance needs to add up to 100%

100-35= 65%

The second isotope will have a 65% abundance.
<span />

7 0

2 years ago

Part III.The two reactions involved in quantitatively determining the amount of iodate in solution are: IO3-(aq) 5 I-(aq) 6 H (a

slega [8]

Answer:

$\large \boxed{\math{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$

Explanation:

The I₂ is the common substance in the two equations.

(1) IO₃⁻ + 5I⁻ + 6H⁺ ⟶ 3I₂ + 3H₂O

{2) I₂ + 2S₂O₃²⁻ ⟶ 2I⁻ + S₄O₆²⁻

From Equation (1), the molar ratio of iodate to iodine is

$\dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{3}{1}$

From Equation (2), the molar ratio of iodine to thiosulfate is

$\dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} = \dfrac{2}{1}$

Combining the two ratios, we get

$\text{Stoichiometric factor} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{IO}_{3}^{-}} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} \times \dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{2}{1} \times \dfrac{3}{1} = \mathbf{\dfrac{6}{1}}\\\\\text{The stoichiometric factor is $\large \boxed{\mathbf{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$}$

7 0

2 years ago

How many sodium ions are in 1.4 kg of sodium chloride, NaCl?

Whitepunk [10]

Answer:

1.44 x 10²⁵ ions of Na⁺

Explanation:

Given parameters:

Mass of NaCl = 1.4kg = 1400g

Unknown:

Number of ions of sodium = ?

Solution:

The compound NaCl in ionic form can be written as;

NaCl → Na⁺ + Cl⁻

In 1 mole of NaCl we have 1 mole of sodium ions

Now, let us find the number of moles in NaCl;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of NaCl = 23 + 35.5 = 58.5g/mol

Number of moles = $\frac{1400}{58.5}$ = 23.93mol

So;

Since 1 mole of NaCl gives 1 mole of Na⁺

In 23.93 mole of NaCl will give 23.93 mole of Na⁺

1 mole of a substance = 6.02 x 10²³ ions of a substance

23.93 mole of a substance = 6.02 x 10²³ x 23.93

= 1.44 x 10²⁵ ions of Na⁺

7 0

2 years ago