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Sladkaya [172]
3 years ago
15

Which of the following atoms most easily loses an electron Ba Cl Te C P

Chemistry
1 answer:
solong [7]3 years ago
7 0
Ba! It is a metal and wants to lose 2 electrons ASAP
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How many grams of calcium cyanide (Ca(CN)2) are contained in 0.79 mol of calcium cyanide?
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I dont know but do you know da wae brudda?


3 0
3 years ago
What is the percentage error of length measurement of 0.229cm if the correct value is 0.225cm
Mars2501 [29]

Percent error is the difference between the measured and known value, divided by the known value, multiplied by 100%.

So first, we take our measured value, .299 cm, minus our known value, .225 cm.

.299 cm - .225 cm=.004 cm

Next, we divide that by our known value

\frac{.004}{.225}=.0177777778

Finally, multiply your answer by 100

.0177777778 x 100= 1.77777778 %

Round to three significant figures, and you're done.

=1.78 % error



5 0
3 years ago
Read 2 more answers
Consider this chemical reaction. What chemical is reduced?<br> CH3OH + NAD --&gt; CH2O + NADH
nikklg [1K]

Answer:

CH3OH and NADH

Explanation:

The given chemical reaction is an redox reaction in which reduction and oxidation take place.

In the process of oxidation: electrons are loss while in the process of reduction: electrons are gained.

In the given redox reaction: CH3OH + NAD --> CH2O + NADH

NAD is reduced to NADH as NADH gains one hydrogen electron while CH3OH (methanol) is oxidized to CH2O (methanal)  by losing electrons.

So, CH3OH (methanol) and NADH are the reduced forms while NAD and CH2O (methanal) are oxidized forms.

5 0
3 years ago
A chemical reaction can theoretically produce 1375.5 grams of product , but in actuality 112.9 grams are product . Which is the
zalisa [80]

Answer:

8.20 % → Percent yield reaction

Explanation:

To find the percent yield of reaction we apply this:

(Produced yield / Theoretical yield) . 100 = %

Produced yield = 112.9 g

Theoretical yield = 1375.5 g

We replace → (112.9g / 1375.5 g) . 100

8.20 % → Percent yield reaction

3 0
3 years ago
10. Isolation of a pure sample of the third product, which has been determined to be an isomer of the major and minor products,
Ymorist [56]

Answer:

Four possible isomers (1–4) for the natural product essramycin. The structure of compound 1 was attributed to essramycin by 1H NMR, 13C NMR, HMBC, HRMS, and IR experiments.

Explanation:

Three synthetic routes were used to prepare all four compounds (Figure 2A). All three reactions utilize 2-(5-amino-4H-1,2,4-triazol-3-yl)-1-phenylethanone (5) as the precursor, whereas each uses different esters (6–8) to construct the pyrimidinone ring. Isomer 1 was prepared by reaction A, which used triazole 5 and ethyl acetoacetate (6) in acetic acid. This was the reaction used in syntheses of essramycin by the Cooper and Moody laboratories.3,4 Reaction B produced compound 2 (minor product) and compound 3 (major product), which were separated chromatographically. This reaction allowed reagent 5 to react with ethyl 3-ethoxy-2-butenoate (7) in the presence of sodium in methanol, under reflux for 24 h. Compound 4 was prepared by reaction C, which was obtained by reflux of 5 and methyl 2-butynoate (8) in n-butanol.

7 0
3 years ago
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