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Vikki [24]
2 years ago
13

The boiling point of a substance is tested. After 10 tests, the result is given as 37+/−3°C. Which conclusion can be drawn from

this result? (1 point)
The scientists do not need to collect more data because they have narrowed down the range of the results.

The scientists should not report these results until they have the exact number.

The actual boiling point is either 34°C or 40°C.

The actual boiling point is probably between 34°C and 40°C.
Chemistry
2 answers:
VMariaS [17]2 years ago
6 0

Answer:

Explanation:

The actual boiling point is probably between 34C and 40C.

sergij07 [2.7K]2 years ago
4 0

C: The actual boiling point is either 34°C or 40°C.

Explanation:

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What is the theoretical yield of NaBr
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The theoretical yield of NaBr given that 2.36 moles of FeBr₃ reacts is 7.08 moles

<h3>Balanced equation </h3>

2FeBr₃ + 3Na₂S → Fе₂S₃ + 6NaBr

From the balanced equation above,

2 moles FeBr₃ reacted to produce 6 moles of NaBr

<h3>How to determine the theoretical yield of NaBr</h3>

From the balanced equation above,

2 moles FeBr₃ reacted to produce 6 moles of NaBr

Therefore,

2.36 moles FeBr₃ will react to produce = (2.36 × 6) / 2 = 7.08 moles of NaBr

Therefore,

Thus, the theoretical yield of NaBr is 7.08 moles

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Answer:

I don't know chemistry

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Unit: Chemical Quantities
Vaselesa [24]

Answer:

(See explanation for further details)

Explanation:

1) The quantity of moles of sulfur is:

n = \frac{1.20\times 10^{24}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }

n = 1.993\,moles

2) The number of atoms of helium is:

x = (1.5\,moles)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)

x = 9.033\times 10^{23}\,atoms

3) The quantity of moles of carbon monoxide is:

n = \frac{4.15\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }

n = 0.689\,moles

4) The number of molecules of sulfur dioxide is:

x = (2.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

x = 1.355\times 10^{24}\,molecules

5) The quantity of moles of sodium chloride is:

n = \frac{2.4\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }

n = 0.399\,moles

6) The number of formula units of magnesium iodide is:

x = (1.8\,moles)\cdot \left(6.022\times 10^{23}\,\frac{f.u.}{mole} \right)

x = 1.084\times 10^{24}\,f.u.

7) The quantity of moles of potassium permanganate is:

n = \frac{3.67\times 10^{23}\,f.u.}{6.022\times 10^{23}\,\frac{f.u.}{mol} }

n = 1.214\,moles

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x = (0.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

x = 1.506\times 10^{23}\,molecules

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n = \frac{3.67\times 10^{23}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }

n = 0.609\,moles

10) The number of molecules of oxygen difluoride is:

x = (3.52\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

x = 2.120\times 10^{24}\,molecules

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