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Troyanec [42]
3 years ago
10

11. A certain gas takes three times as long to effuse out

Chemistry
1 answer:
gladu [14]3 years ago
8 0

answer is 36 u 2 one okkkkkkk

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Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction: P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH
Pie

Answer: -1835 kJ

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)  \Delta H_1=+157kJ (1)

P_4(g)+6Cl_2\rightarrow 4PCl_3(g)  \Delta H_2=-1207J (2)

Net chemical equation:

P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)  \Delta H=? (3)

Multiplying equation (1) by 4,  and reversing we get

4PCl_3(g)+4Cl_2(g)\rightarrow 4PCl_5(s)  \Delta H_4=4\times -157kJ=-628kJ (4)

Adding (2) and (4)

P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)  \Delta H_3=\Delta H_2+\Delta H_4=-1207kJ-628kJ=-1835kJ  

Thus enthaply change for the reaction is -1835 kJ.

3 0
4 years ago
How many moles is<br> 55 g of Naci?<br> 0.025 g of NaCO3?
mart [117]

Answer:

0.940mol &

0.000301mol respectively.

Explanation:

number of moles = given mass / molar mass

given mass of Nacl = 55g Molar mass = 23 + 35.5

n=m/M = 55g/58.5g/mol = 0.940mol

note- (add the atomic weights of sodium and chlorine to get the molar mass of Nacl.) = 58.5g/mol

similarly, NaCO3 = 23 + 12 + 16*3 = 83g/mol

n=m/M = 0.025g/83g/mol = 3.01 * 10^-4 = 0.000301mol

extra: If you ever get asked to put it in number of particles just use the relation of 1mole = 6.02 * 10^23 particles.

8 0
3 years ago
PLZ SOMEONE HELP THANKS:))))<br> materials and conclude which is best to use. Full explanation
grin007 [14]

Answer:

For quick facts use reference materials such as encyclopedias, almanacs, or dictionaries. For credible and reputable information, use online databases of academic journals, or peer-reviewed journal articles that are written by scholars and researchers for a specific discipline.

3 0
3 years ago
In which image is the wave amplitude the greatest?<br> 4<br> 3<br> 2<br> 1
gladu [14]
Answer:4
Step by step
7 0
3 years ago
What mass of lithium phosphate would you mass to make 2.5 liter of 1.06 M lithium
alex41 [277]

Answer:

Approximately 3.06 \times 10^{2}\; \rm g (approximately 306\; \rm g.)

Explanation:

Calculate the quantity n of lithium phosphate in V = 2.5\; \rm L of thisc = 1.06\; \rm M = 1.06\; \rm mol \cdot L^{-1} lithium phosphate solution.

\begin{aligned}n &= c \cdot V\\ &= 2.5\; \rm L \times 1.06\; mol \cdot L^{-1}\\ &= 2.65\; \rm mol\end{aligned}.

Empirical formula of lithium phosphate: \rm Li_3PO_4.

Look up the relative atomic mass of \rm Li, \rm P,and \rm O on a modern periodic table:

  • \rm Li: 6.94.
  • \rm P: 30.974.
  • \rm O: 15.999.

Calculate the formula mass of \rm Li_3PO_4:

M(\rm Li_3PO_4) = 3 \times 6.94 + 30.974 + 4 \times 15.999 = 115.79\; \rm g \cdot mol^{-1}.

Calculate the mass of that n = 2.65\; \rm mol of \rm Li_3PO_4 formula units:

\begin{aligned}m &= n \cdot M \\ &= 2.65\; \rm mol \times 115.79\; \rm g\cdot mol^{-1} \\ &\approx 3.06 \times 10^{2}\; \rm g \end{aligned}.

5 0
3 years ago
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