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Archy [21]
3 years ago
6

Approximate the instantaneous rate of this reaction at time t = 40 s.

Chemistry
1 answer:
Vadim26 [7]3 years ago
6 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Find the slope at t = 40, points (0, 0.48) and (70, 0.24).
Slope= Δy= (0.24 – 0.48)=– 0.0034 M/s
            Δx(70 – 0)
<span>
(the negative represent direction so the answer is positive)</span>
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Electron A falls from energy level X to energy level Y and releases blue light. Electron B falls from energy level Y to energy l
MrMuchimi

Answer: Transition from X to Y will have greater energy difference.

Explanation: For studying the energy difference, we require Planck's equation.

                                E=\frac{hc}{\lambda}

where, h = Planck's Constant

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5 0
3 years ago
In which medium does sound travel the slowest?
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3 years ago
Read 2 more answers
HELPP
slamgirl [31]

Answer:

In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.

Explanation:

tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=

mass compound

mass H

×100%

\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=

mass compound

mass C

×100%

If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=

10.0g compound

2.5g H

×100%=25%

\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=

10.0g compound

7.5g C

×100%=75%

7 0
3 years ago
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