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OleMash [197]
3 years ago
8

A chemistry student’s height is measured at 68.5 inches. How tall is the student in cm?

Chemistry
1 answer:
pantera1 [17]3 years ago
4 0
173.99cm multiply 68.5 x 2.54 always use that number when converting inches to cm
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Helpp plssssss :)<br> .....
MaRussiya [10]
B is right yuh yuh yuh yuh
4 0
3 years ago
How many grams of H2SO4 are needed to prepare 500. mL of a .250M solution?
zavuch27 [327]

Answer:

We need 12.26 grams H2SO4

Explanation:

Step 1: Data given

Volume of a H2SO4 solution = 500 mL = 0.500 L

Concentration of the H2SO4 solution = 0.250 M

Molar mass of H2SO4 = 98.08 g/mol

Step 2: Calculate moles H2SO4

Moles H2SO4 = concentration * volume

Moles H2SO4 = 0.250 M * 0.500 L

Moles H2SO4 = 0.125 moles

Step 3: Calculate mass of H2SO4

Mass of H2SO4 = moles * molar mass

Mass of H2SO4 = 0.125 moles * 98.08 g/mol

Mass of H2SO4 = 12.26 grams

We need 12.26 grams H2SO4

7 0
3 years ago
What is the molarity of a 799 mL solution that contains 3.3 moles of NaNO3?
rusak2 [61]

Answer:   I think It might be 1 M???

Explanation:  Sorry I'm not in high school I put the wrong age

8 0
3 years ago
N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol
docker41 [41]

Answer:

A

Explanation:

Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:


\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)

Therefore, from the chemical equation, we have that:


\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} +  \Delta H^\circ_f \text{ H$_2$O}  \right]   -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}

In conclusion, our answer is A.

5 0
2 years ago
the frequency of a given region of the electromagnetic spectrum is more than 3 x 10^19HZ. note that the speed of light is 2.998
Cerrena [4.2K]
<span>To solve this problem, You need to look up a picture/diagram of the electromagnetic spectrum. This will have the wave regions listed as well</span> as frequencies and wavelength.
Wavelength is distance/length of one wave, which can be calculated using frequency (hz = s^-1) and the speed of light.
2.998 x 10^8 m/s ÷ 3 x 10^19 s^-1 = 9.99 x 10^-12 m

The Frequency given falls in between X-rays and Gamma rays. The wavelength however; is in the Gama ray region.




5 0
3 years ago
Read 2 more answers
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