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3 years ago

7

P4(s) + O2(g) → __ P4O10(s) To balance this reaction equation, first write correct formulas for the reactants and products

, leaving the stoichiometric coefficients blank. We will balance P first. How many P4 molecules are needed to balance P for each P4O10?

1 answer:

Vedmedyk [2.9K]3 years ago

8 0

Answer:

P4(s) + 5 O2 (g)→ P4O10

Explanation:

If we desire to write a balanced chemical reaction equation, the rule of thumb is simple; the number of atoms of each element on the right hand side of the reaction equation must be the same as the number of atoms of the same element on the left hand side of the reaction equation. Once this condition is satisfied, the reaction equation is said to be balanced.

As we can see, we need one mole of P4 and five moles of O2 to produce one mole of P4O10.

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The diagram below shows part of the rock cycle. (6 points)

Anna35 [415]

Answer:

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Assuming this is a cycle, the volcanic eruption would lead back to rock B; rocks formed by volcanic eruptions are considered Igneous.

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Highly liquid lava forms wide, shield-like mountains. True False

timurjin [86]

This answer is True :)

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The half-life of a radioactive isotope is the amount of time it takes for a quantity of that isotope to decay to one half of its

Sedaia [141]

Radio active decay reactions follow first order rate kinetics.

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$t_{\frac{1}{2}} =$ $\frac{ln 2}{k}$

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Where k is the decay constant

b) Finding out the decay constant for the decay of C-14 isotope:

$Decay constant (k) = \frac{0.693}{t_{\frac{1}{2}}}$

$k = \frac{0.693}{5230 years}$

$k = 1.325 * 10^{-4} yr^{-1}$

c) Finding the age of the sample :

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The first order rate equation is,

$[A] = [A_{0}]e^{-kt}$

$\frac{[A]}{[A_{0}]} = e^{-kt}$

$\frac{35}{100} = e^{-(1.325 *10^{-4})t}$

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3 0

3 years ago

The solubility of magnesium phosphate at a given temperature is 0.173 g/L. Calculate the Ksp at this temperature. After you calc

jek_recluse [69]

Answer: $K_{sp}=1.25\times 10^{-14}$

$pK_{sp}=13.90$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation for the ionization of magnesium phosphate is given as:

$Mg_3(PO_4)_2\rightarrow 3Mg^{2+}+2PO_4^{3-}$

When the solubility of $Mg_3(PO_4)_2$ is S moles/liter, then the solubility of $Mg^{2+}$ will be 3S moles\liter and solubility of $PO_4^{3-}$ will be 2S moles/liter.

Thus S = 0.173 g/L or $\frac{0.173g/L}{262.8g/mol}=0.00065mol/L$

$K_{sp}=(3S)^3\times (2S)^2$

$K_{sp}=108S^5$

$K_{sp}=108\times (0.00065)^5=1.25\times 10^{-14}$

$pK_{sp}=-log(K_{sp})=\log (1.25\times 10^{-14})=13.90$

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2 years ago

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Ber [7]

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