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ELEN [110]
3 years ago
9

Balanced equation for calcium carbonate and hydrobromic acid

Chemistry
1 answer:
Luda [366]3 years ago
3 0

Answer:

2 HBr (aq) + CaCO3 (s) → CaBr2 (aq) + CO2 (g) + H2O (l)

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Which feature of a hypothesis leads to a new experiment
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Find the mass of 3.27 x 10^23 molecules of H2SO4. Use 3 significant digits<br> and put the units.
marta [7]

Answer:

Approximately 53.3\; \rm g.

Explanation:

Lookup Avogadro's Number: N_{\rm A} = 6.02\times 10^{23}\; \rm mol^{-1} (three significant figures.)

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  • \rm H: 1.008.
  • \rm S: 32.06.
  • \rm O: 15.999.

(For example, the relative atomic mass of \rm H is 1.008 means that the mass of one mole of \rm H\! atoms would be approximately 1.008\! grams on average.)

The question counted the number of \rm H_2SO_4 molecules without using any unit. Avogadro's Number N_{\rm A} helps convert the unit of that count to moles.

Each mole of \rm H_2SO_4 molecules includes exactly (1\; {\rm mol} \times N_\text{A}) \approx 6.02\times 10^{23} of these \rm H_2SO_4 \! molecules.

3.27 \times 10^{23} \rm H_2SO_4 molecules would correspond to \displaystyle n = \frac{N}{N_{\rm A}} \approx \frac{3.27 \times 10^{23}}{6.02 \times 10^{23}\; \rm mol^{-1}} \approx 0.541389\; \rm mol of such molecules.

(Keep more significant figures than required during intermediary steps.)

The formula mass of \rm H_2SO_4 gives the mass of each mole of \rm H_2SO_4\! molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:

\begin{aligned}& M({\rm H_2SO_4}) \\ &= (2 \times 1.008 + 32.06 + 4 \times 15.999)\; \rm g \cdot mol^{-1} \\ &= 98.702\; \rm g \cdot mol^{-1}\end{aligned}.

Calculate the mass of approximately 0.541389\; \rm mol of \rm H_2SO_4:

\begin{aligned}m &= n \cdot M \\ &\approx 0.541389\; \rm mol \times 98.702\; \rm g \cdot mol^{-1}\\ &\approx 53.3\; \rm g\end{aligned}.

(Rounded to three significant figures.)

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What is the half-life of an isotope that decays to 12.5% of its original activity in 19.8 hours?
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x= x_{0}  e^{kt} ,where~ x ~is~ the ~ |amount\ of \material\ at\ any~~time~t&#10;
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when~x=  \frac{12.5}{100} x_{0}  }= \frac{125}{1000}  x_{0}
\frac{x}{ x_{0} } = \frac{125}{1000}= \frac{1}{8}
when t=19.8 hrs,&#10;x= x_{0}  e^{19.8 k}, &#10; \frac{x}{ x_{0} } = e^{19.8 k}, &#10; \frac{1}{8} = e^{19.8k} ,&#10;ln \frac{1}{8} =19.8 k,&#10;
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