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Rainbow [258]
3 years ago
12

What labels in the areas marked Y and Z

Chemistry
1 answer:
MrMuchimi3 years ago
3 0

Answer:

Can you please provide a graph or picture?

Explanation:

.

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What is/are the principal product(s) formed when excess methylmagnesium iodide reacts with p-hydroxyacetophenone?
Montano1993 [528]

The end product will depend upon

a) the amount of the reagent taken

b) the final treatment of the reaction

If we have just taken methylmagnesium iodide and p-hydroxyacetophenone, then we will get methane and hydroxyl group substituted with MgI in place of hydrogen

Figure 1

However if we have taken excess of methylmagnesium iodide which is Grignard's reagent followed by hydrolysis we will get different product

Figure 2

5 0
3 years ago
Which list of elements consists of a metal, a metalloid, and a nonmetal?
Irina18 [472]

Answer:

Sn, Si, C

Explanation:

8 0
3 years ago
La masa molar para el siguiente compuesto: CuOH, si sus masas atómicas son Cu=64 g/mol; O=16 g/mo e H=1 g/mol
aev [14]

Answer:

Abraham Lincoln

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4 0
3 years ago
Read 2 more answers
What is the wavelength if the frequency is 29.2 hz?
ZanzabumX [31]

Answer : The wavelength is 1.027\times 10^7m

Solution : Given,

frequency = 29.2 Hz

Formula used :

\nu=\frac{c}{\lambda}\\\lambda=\frac{c}{\nu}

where,

\nu = frequency

\lambda = wavelength

c = speed of light = 3\times 10^8m/s

Now put all the given values in this formula, we get the wavelength.

\lambda=\frac{3\times 10^8m/s}{29.2Hz}=0.1027\times 10^8m=1.027\times 10^7m                 (1Hz=1s^{-1})

Therefore, the wavelength is 1.027\times 10^7m

8 0
3 years ago
a solution with a transmittance of 0.44 is analyzed in a spectrophotometer with 6% stray light. calculate the absorbance reporte
IgorLugansk [536]

The absorbance reported by the defective instrument was 0.3933.

Absorbance A = - log₁₀ T

Tm = transmittance measured by spectrophotometer

Tm = 0.44

Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654

True absorbance can be calculated by true transmittance, Tm = T+S(α-T)

S = fraction of stray light = 6%= 6/100 = 0.06

α= 1, ideal case

T = true transmittance of the sample

Tm = T+S(α-T)

now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233

therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)

i.e; 0.3933

To know more about transmittance click here:

brainly.com/question/17088180

#SPJ4

3 0
1 year ago
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