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eduard
3 years ago
6

Why does the half-life of cancer treatment decay?

Chemistry
1 answer:
zimovet [89]3 years ago
3 0

Answer:

The decay of radioactive elements occurs at a fixed rate. The half-life of a radioisotope is the time required for one half of the amount of unstable material to degrade into a more stable material. For example, a source will have an intensity of 100% when new. At one half-life, its intensity will be cut into 50% of the original intensity.

Explanation:

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Consider the reaction of magnesium metal with hydrochloric acid to produce magnesium chloride and hydrogen gas. If 3.29 mol of m
gregori [183]

Answer:

1.645 moles of excess reactant that is of magnesium metal are left over.

Explanation:

Moles of magnesium metal = 3.29 mol

Moles of HCl = 3.29 mol

Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)

According to recation, 2 moles of HCl reacts with 1 mol of magnesium metal, then 3.29 moles of HCl will react with :

\frac{1}{2}\times 3.29 mol=1.645 mol of magnesium metal

Moles of HCl left = 3.29mol - 3.29 mol = 0

Moles of magnesium metal left = 3.29 mol - 1.645 mol = 1.645 mol

1.645 moles of excess reactant that is of magnesium metal are left over.

7 0
3 years ago
Pitchblende is ore of _____.
Valentin [98]
Its an ore of uraninte i think.
6 0
4 years ago
Read 2 more answers
Calculate the mass (in grams) of 250mL of ether at 25 oC. The density of
leonid [27]
  • Volume=250mL
  • Density=0.71g/ml

\boxed{\sf Density=\dfrac{Mass}{Volume}}

\\ \sf{:}\implies Mass=Density(Volume)

\\ \sf{:}\implies Mass=0.71(250)

\\ \sf{:}\implies Mass=177.5g

6 0
3 years ago
The image shows a portion of the periodic table. Which two elements have similar characteristics?
tatuchka [14]

Answer:

C

Explanation:

Similar characteristics depend on columns.

Flourine and Chlorine are both halogens so they will have similar properties.

5 0
3 years ago
Read 2 more answers
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
3 years ago
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