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Burka [1]
3 years ago
14

An unknown substance is a yellow gas at room temperature. Is this substance ionic or covalent?

Chemistry
2 answers:
Free_Kalibri [48]3 years ago
8 0

Answer:

Iconic

Explanation:

An ionic bond is between a metal and a nonmetal, and a covalent bond is between 2 nonmetals.

DIA [1.3K]3 years ago
7 0
Covalent because it is yellow maybe
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Acetone (fingernail-polish remover) has a density of 0.7857 g/cm3 What is the mass in grams of 17.16 mL of acetone?
Mashutka [201]
1 mL = 1 cm³

D = m / V

0.7857 = m / 17.16

m = 0.7857 x 17.16

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8 0
3 years ago
how many ml of a 22.5% (v/v) ethanol solution would you need to measure out in order to have 12.5 ml of ethanol ?
Aleks [24]
Volume percent<span> or </span>volume/volume percent<span> (v/v%) is used when preparing solutions of liquids. It will have units of volume of the smaller composition substance over the volume of the solution. We calculate as follows:

12.5 mL ethanol = .225 mL ethanol / 1 mL solution ( V )
V = 55.56 mL of the 22.5 % by volume ethanol solution is needed

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8 0
4 years ago
Some organisms that are alive today do not look exactly like organisms that lived in the past. However, if a comparison of DNA s
ololo11 [35]

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5 0
4 years ago
Read 2 more answers
Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m
erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3}  =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2}  =0.06882molBr^-

Now, we compute the total moles of bromide:

n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol

Then, the total volume in liters:

150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L

Therefore, the concentration of total bromide is:

M=\frac{0.12338mol}{0.325L}\\\\M=0.380M

Best regards!

8 0
3 years ago
Tris {(hoch2)3cnh2} is one of the most common buffers used in biochemistry. a solution is prepared by adding enough tris and 12
shusha [124]

Given that,

The concentration of TRIS = 0.30 M

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Kb = 1.2 x 10^-6

pKb = -log Kb = - log (1.2 x 10^-6) = 5.920

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<span>pOH=14-pH=14-6.221 = 7.779</span>

7 0
3 years ago
Read 2 more answers
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