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7 months ago

15

Choose the chemical equation that is correctly balanced. 2ca(s) cl2(g) → cacl2(s) 4mg(s) o2(g) → 2mgo(s) li(s) cl2(g) → 2licl(s)

c(s) o2(g) → co2(g)

1 answer:

Nadusha1986 [10]7 months ago

8 0

The final one because there is 1 C on each side and 2 O

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Write the symbol of magnesium and iron?

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What volume in mL of 0.220 M HBr solution is required to produce 0.0130 moles of HBr?

yulyashka [42]

Answer: A volume of 59 mL of 0.220 M HBr solution is required to produce 0.0130 moles of HBr.

Explanation:

Given: Moles = 0.0130 mol

Molarity = 0.220 M

Molarity is the number of moles of solute present in liter of a solution.

$Molarity = \frac{moles}{volume (in L)}$

Substitute the values into above formula as follows.

$Molarity = \frac{moles}{volume (in L)}\\0.220 M = \frac{0.0130 mol}{Volume (in L)}\\Volume (in L) = 0.059 L$

As 1 L = 1000 mL

So, 0.059 L = 59 mL

Thus, we can conclude that a volume of 59 mL of 0.220 M HBr solution is required to produce 0.0130 moles of HBr.

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2 years ago

) B5H9(l) is a colorless liquid that will explode when exposed to oxygen. How much heat is released when 0.211 mol of B5H9 react

Tom [10]

<u>Answer:</u> The amount of heat released when 0.211 moles of $B_5H_9(l)$ reacts is 554.8 kJ

<u>Explanation:</u>

The chemical equation for the reaction of $B_5H_9$ with oxygen gas follows:

$2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(5\times \Delta H_f_{(B_2O_3(s))})+(9\times \Delta H_f_{(H_2O(l))})]-[(2\times \Delta H_f_{(B_5H_9(l))})+(12\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.4kJ/mol\\\Delta H_f_{(B_2O_3(s))}=-1272kJ/mol\\\Delta H_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2\times (-1272))+(9\times (-285.4))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H_{rxn}=-5259kJ$

To calculate the amount of heat released for the given amount of $B_5H_9(l)$ , we use unitary method, we get:

When 2 moles of $B_5H_9(l)$ reacts, the amount of heat released is 5259 kJ

So, when 0.211 moles of $B_5H_9(l)$ will react, the amount of heat released will be = $\frac{5259}{2}\times 0.211=554.8kJ$

Hence, the amount of heat released when 0.211 moles of $B_5H_9(l)$ reacts is 554.8 kJ

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2 years ago

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Answer:

B

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as temperature rises, the particles gain kinetic energy (they will move faster) so if temperatures stays constant, so will the movement or vibration of the particles

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