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vesna_86 [32]
3 years ago
14

The set of norms that establishes and characterizes the relationship between married individuals.

Arts
2 answers:
eimsori [14]3 years ago
6 0

Answer: marriage

Explanation:

Jet001 [13]3 years ago
5 0

Answer:

marriage

Explanation:

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Is “Four Minutes Thirty-Three Seconds” by John Cage really music? Why or why not?
kobusy [5.1K]

Answer:

No, because he was just sitting there instead of playing. So it's called silence instead of music.

Explanation:

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3 years ago
If one scans topics and pages of Leonardo da Vinci's notebooks, it becomes apparent that he
Alex787 [66]
It would appear that he was very secretive of his work , it was even thought that da vinci wrote his work backwards that way it would become harder to read . It would be d . strictly for his own use . 
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3 years ago
Bianca has to prepare an e-learning module. Which tool will help her collect information from different multimedia sources and o
Tcecarenko [31]

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D

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I hope this helps!

4 0
3 years ago
Describe the preparation of 2.00 l of 0.100 m glycine buffer, ph 9.0, from glycine and 1.00 m naoh. what mass of glycine is requ
Ksivusya [100]
<span>pH = pKa + log ([R-]/[RH]) Where pH is the pH of the buffer, [R-] is the concentration of the basic species, and [RH] is the concentration of the acidic species. At pH 2.4, the amino group on glycine (pKa = 9.6) will be, for accounts and purposes, 100% protonated. This means our buffer will be dealing with the two ionic forms of the carboxyl group (pKa = 2.4). When pH = pKa, the two species are in equilibrium. This can be seen using the HH equation: 2.4 = 2.4 + log ([R-]/[RH]) 0 = log ([R-]/[RH]) 1 = ([R-]/[RH]) [RH] = [R-] Now we add in another equation, our conservation of mass. M = [RH] + [R-] where M is the molarity of the buffer But since [RH] = [R-]: M = 2 [RH] 0.2 = 2 [RH] And we wind up with: [RH] = [R-] = 0.1 M Now to figure out the moles of each needed, we multiply by the volume of the buffer. 0.1 M * 0.1 L = 0.01 mol This shows that to make 100 ml of 0.2 M glycine buffer, we'll need 0.01 mol of each species. 0.01 mol of 0.5 M HCl: 0.5 mol HCl / 1 L = 0.01 mol / v solve for v v/1 = 0.01 / 0.5 ==> v = 0.02 L or 20 mL weight of glycine: MW: 75.07 g/mol 0.01 mol glycine * (75.07g glycine / 1 mol) = 0.75 g glycine And there's your answer -------- To make this buffer you would add 0.75g glycine to 20 mL of 0.5 M HCl and fill with water until a 100mL volume was achieved.</span>
8 0
4 years ago
Lupe had a difficult time during her last group discussion. She knew her topic, but she got confused as others were talking, esp
Yuri [45]
She should listen for a connection she can make
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