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romanna [79]
3 years ago
9

On Darwin's tree of life, organisms at the base of

Chemistry
2 answers:
bazaltina [42]3 years ago
7 0
The correct answer is part A
masya89 [10]3 years ago
3 0
B - Evolve into the closest species on the tree’s trunk
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Help, please? will mark brainliest​
alexira [117]

Answer:

Sulfur would gain electrons

Explanation:

Atoms want to have a complete out valence shell and because sulfur only needs 2 more electrons to complete the outer shell it would take 2 more.

8 0
3 years ago
Convert 146 calories to kilojoules need to know ASAP
Vikentia [17]
610.864 Energy result in kilojoules:
5 0
3 years ago
What volume is needed to make a 2.45 M solution of KCl using 0.50 mol of KCl
Ivan

Answer:

Explanation:

91.4

grams

Explanation:

C

=

m

o

l

v

o

l

u

m

e

2.45

M

=

m

o

l

0.5

L

2.45

M

⋅

0.5

L

=

m

o

l

m

o

l

=

1.225

Convert no. of moles to grams using the atomic mass of K + Cl

1.225

m

o

l

⋅

(

39.1

+

35.5

)

g

m

o

l

1.225

m

o

l

⋅

74.6

g

m

o

l

=

1.225

⋅

74.6

g

=

91.4

g

4 0
3 years ago
QUESTION 1 Species diversity is comprised of O Composition and Richness O Richness and Frequency O Evenness and Composition Rich
34kurt

Answer:

The correct option is: Richness and Evenness

Explanation:

Species diversity is the measurement of the biological diversity and relative abundance of each species found in a particular biological community.

It is comprised of the following three components:

1. richness (number of species)

2. phylogenetic diversity

3. evenness (relative abundance of each species)

3 0
3 years ago
At 250°C an equilibrium mixture of SbCl3(g), Cl2(g), and SbCl5(g) has the partial pressures 0.670 bar, 0.438 bar, and 0.228 bar,
Sveta_85 [38]

Explanation:

As the given reaction will be SbCl_{3}(g) + Cl_{2}(g) \rightarrow SbCl_{5}

Since, it is given that volume of reaction vessel is doubled. Hence, moles of species involved will also increase.

Therefore, the reaction equation will become as follows.

             2SbCl_{3}(g) + 2Cl_{2}(g) \rightarrow 2SbCl_{5}

As it is given that partial pressure of SbCl_{3} is 0.670 bar, Cl_{2} is 0.438 bar and SbCl_{5} is 0.228 bar.

Expression to calculate new equilibrium pressure if the volume of reaction vessel is double is as follows.

                  K_{p} = \frac{[P_{SbCl_{5}}]^{2}}{[P_{SbCl_{3}}]^{2}[P_{Cl_{2}}]^{2}}

                                   = \frac{(0.228)^{2}}{(0.670)^{2}(0.438)^{2}}

                                   = 6.05 \times 10^{3}

Thus, we can conclude that new equilibrium pressures if the volume of the reaction vessel is doubled is 6.05 \times 10^{3}.

4 0
4 years ago
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