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4 years ago

Explain the relationships among balancing, chemical equations, and endothermic/exothermic

Chemistry

1 answer:

Ket [755]4 years ago

3 0

Endothermic - absorbs heat from the surroundings
exothermic - releases heat into the surroundings

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What is the empirical formula of a compound composed of 3.25% hydrogen ( H ), 19.36% carbon ( C ), and 77.39% oxygen ( O ) by ma

Montano1993 [528]

Answer:

The answer to your question is C₂HO₃

Explanation:

Data

Hydrogen = 3.25%

Carbon = 19.36%

Oxygen = 77.39%

Process

1.- Write the percent as grams

Hydrogen = 3.25 g

Carbon = 19.36 g

Oxygen = 77.39 g

2.- Convert the grams to moles

1 g of H ----------------- 1 mol

3,25 g of H ------------- x

x = (3.25 x 1) / 1

x = 3.25 moles

12 g of C ---------------- 1 mol

19.36 g of C ---------- x

x = (19.36 x 1) / 12

x = 1.61 moles

16g of O --------------- 1 mol

77.39 g of O --------- x

x = (77.39 x 1)/16

x = 4.83

3.- Divide by the lowest number of moles

Carbon = 3.25/1.61 = 2

Hydrogen = 1.61/1.61 = 1

Oxygen = 4.83/1.61 = 3

4.- Write the empirical formula

C₂HO₃

4 0

3 years ago

How many liters of oxygen are in 8.32 moles of oxygen at STP

Jlenok [28]

Answer:

186 Liters at STP conditions

Explanation:

1 mole of any gas at STP conditions occupies 22.4 Liters.

Therefore, 8.32 moles O₂(g) = 8.32 moles x 22.4Liters/mole = 186 Liters (3 sig.figs.)

6 0

3 years ago

Amanda [17]

The Kinitec Molecular Theory of Matter
Yes Kninetic energy

5 0

4 years ago

in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc

Margarita [4]

Answer: 0.1161 grams of mercury(II) sulfide) form.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$ .....(1)

a) Molarity of $Na_2S$ solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

$0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol$

$\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol$

b) Molarity of $Hg(NO_3)_2$ solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

$0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol$

$Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3$

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ reacts with 1 mole of $Na_2S$

Thus 0.0005 moles of $HgNO_3$ reacts with= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg(NO_3)_2$

Thus $Hg(NO_3)_2$ is the limiting reagent and $Na_2S$ is the excess reagent.

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ forms= 1 mole of $Hg_2S$

Thus 0.0005 moles of $Hg(NO_3)_2$ forms= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg_2S$

mass of $H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g$

Thus 0.1161 grams of mercury(II) sulfide) form.

5 0

3 years ago

The cell membrane responds to chemical signals from other cells A) true B) false

nataly862011 [7]

The answer would be A)true

8 0

4 years ago

Read 2 more answers