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Helen [10]
3 years ago
7

What can be a catchphrase for cobalt if it were a superhero?

Chemistry
1 answer:
creativ13 [48]3 years ago
7 0

Answer:lumerman.com › catchphrases › catchphrase

Welcome true believers! The first step to being a superhero is to have a superhero catchphrase to yell out ... I am Danger Boy, the Mighty, Loud, Confidant of Might!!! 02/23/21 ... I am cobalt, the Dynamic, Unbeatable, Hero of Hono

Explanation:

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Can someone PLEASE HELP me? <br><br> How can you tell if a bond is Covalent?
g100num [7]
There is a couple different ways to determine if a bond is ionic orcovalent. By definition, an ionicbond is between a metal and a nonmetal, and a covalent bond is between 2 nonmetals. So you usually just look at the periodic table and determine whether your compound is made of a metal/nonmetal or is just 2 nonmetals.
6 0
3 years ago
Which of the following statements apply to chemistry?
Maksim231197 [3]
chemistry is used in medicine and water and air are chemical
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Read 2 more answers
Calcuate the number of<br> grams of solute in 453.9mL<br> of 0.237 M calcium acetate
AysviL [449]

The number of  grams : 17.082 g

<h3>Further explanation</h3>

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

\large {\boxed {\bold {M ~ = ~ \dfrac {n} {V}}}

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

453.9 mL  of 0.237 M calcium acetate

  • mol

\tt mol=M\times V=0.237\times 0.4539=0.108

  • mass

MW Ca(C₂H₃OO)₂ : 158,17 g/mol

\tt mass=mol\times MW\\\\mass=0.108\times 158.17=17.082~g

4 0
2 years ago
A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained. How many
beks73 [17]

<u>Answer: </u>

A sample initially contained 150 mg of radon-222. After 11.4 days only 18.75mg of the radon-222 in the sample remained where 3 half-lives have passed

<u>Explanation:</u>

Given, the initial value of the sample, A_0 = 150mg

Final value of the sample or the quantity left, A = 18.75mg

Time = 11.4 days

The amount left after first half life will be ½.

The number of half-life is calculated by the formula

\frac{A}{A_{0}}=\left(\frac{1}{2}\right)^{N}

where N is the no. of half life

Substituting the values,

\frac{18.75}{150}=\left(\frac{1}{2}\right)^{N}

\left(\frac{1}{2}\right)^{3}=\left(\frac{1}{2}\right)^{N}

On equating, we get, N = 3

Therefore, 3 half-lives have passed.

7 0
3 years ago
The vapor pressure of benzene is 73.03 mm Hg at 25°C. How many grams of estrogen (estradiol), C18H24O2, a nonvolatile, nonelectr
alexira [117]

Answer:

14.9802 grams of estrogen must be added to 216.7 grams of benzene.

Explanation:

The relative lowering of vapor pressure of solution containing non volatile solute is equal to mole fraction of solute.

\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}

Where:

p_o = Vapor pressure of pure solvent

p_s = Vapor pressure of the solution

n_1 = Number of moles of solvent

n_2 = Number of moles of solute

p_o = 73.03 mmHg

p_s= 71.61 mmHg

n_1=\frac{216.7 g}{78.12 g/mol}=2.7739 mol

\frac{73.03 mmHg-71.61 mmHg}{73.03 mmHg}=\frac{n_2}{2.7739 mol+n_2}

n_2=0.05499 mol

Mass of 0.05499 moles of estrogen :

= 0.05499 mol × 272.4 g/mol = 14.9802 g

14.9802 grams of estrogen must be added to 216.7 grams of benzene.

6 0
2 years ago
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