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2 years ago

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Identify the type of reaction, write the products with phases, and balance. Show all work.

1 answer:

iren [92.7K]2 years ago

8 0

The correct classification of the reactions are as follows:

<span>1) 2Mg(s) + O2(g)----> 2MgO (s)
It is a synthesis reaction or a combination reaction. It involves the bonding of the reactant molecules forming one single molecule with new set of properties.

2) Na(s) + Cl2(g)------> NaCl(s)
It is a combination reaction as well, the same as the first reaction.

3). PbCO3(s)------> PbO(s)+Co2(g)
It is a decomposition reaction where a single molecule is breaks down into multiple molecules.

4). Ch4(g) + 2O2(g)----> CO2(g) + 2H2O(g)
It is a combustion reaction since it involves the reaction with oxygen and the production od carbon dioxide and water.</span>

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2 years ago

One mole of carbon, sulfur, and aluminum, respectively. Which jar contains the most atoms?

forsale [732]

Answer:

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Explanation:

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2 years ago

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In an experiment, 107.9 grams of H2SO, is produced when 196.2 grams

Ipatiy [6.2K]

Answer:

54.99% yield

Explanation:

percent yield is just the amount you obtained over the amount expected times 100%.

(experimental value/theoretical value) x 100%

= (107.9 g/196.2 g) x 100%

=54.99% yield

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2 years ago

What is the effect of dilution on the pH of a buffer?

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2 years ago

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When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t

OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

$Mn(s)+2HCl(aq)\rightarrow MnCl_2(aq)+H_2(g)$

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

$m=1.00 g/mL\times 100.0 mL = 100 g$

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

$q=m\times c\times (T_{final}-T_{initial})$

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = $4.18 J/^oC$

$T_{final}$ = final temperature = $23.1^oC$

$T_{initial}$ = initial temperature = $28.9^oC$

Now put all the given values in the above formula, we get:

$q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC$

$q=2,242.4 J=2.242 kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = $\frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol$

$\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol$

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

8 0

2 years ago