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Tresset [83]
3 years ago
10

When comparing the colors of the following four compounds, which is most likely to appear green? [Co(CN)6]3− [Co(H2O)6]3+ [Co(en

)3]3+ [CoI6]3− and why?
Chemistry
2 answers:
Wewaii [24]3 years ago
8 0

Answer:

[Co(H2O)6]3 is most likely to appear green,because of it physical nature and ph values.

alexandr1967 [171]3 years ago
5 0

Answer:

3

Explanation:

its bc

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What is the bond order for a single covalent bond?​
hram777 [196]

Answer:

We typically represent covalent bonds with a dash ( - ) between the atoms. This indicates a single bond. Ex: Cl - Cl

Single bond, double bond, triple bond.

Explanation:

We call it a single covalent bond because the atoms are sharing a single pair of electrons.

4 0
3 years ago
The part of the flower responsible for producing pollen (sperm) is the _______.
IgorC [24]
Hello,

Here is your answer:

The proper answer to this question is option C "stigma".

Here is how:

The stigma is responsible for producing pollen in a plant.

Your answer is C.

If you need anymore help feel free to ask me!

Hope this helps!
7 0
3 years ago
What is the volume of 120mg of water
iVinArrow [24]

Answer:

560mg

Explanation:

120mg-560mg trust dude

7 0
3 years ago
Elements ending in the electron configurations ns^1 are highly reactive metals . What family does these elements belong to ?
myrzilka [38]

Answer:

<h3>A . Alkali metals</h3>

Explanation:

The highlighted elements of the periodic table belong to the alkali metal element family. The alkali metals are recognized as a group and family of elements. These elements are metals. Sodium and potassium are examples of elements in this family.

hope this helps

6 0
3 years ago
What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?
Sergeu [11.5K]

Answer:

\large \boxed{\text{b. 1.5 h}}

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}

2. Calculate the half-life

t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463  h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}

4 0
3 years ago
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