Categorize the following reaction: C3H8 + O2 ---> CO2 + H2O *

When acids react with water, ions are released which then combine with water molecules to form .

mestny [16]

When acids react with water, H ions are released which then combine with water molecules to form H₃O⁺

7 0

4 years ago

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At STP how many moles or helium would occupy a volume of 12 liters?

butalik [34]

1 mole ------------- 22.4 L ( at STP )
?? mole ---------- 12 L

12 x 1 / 22.4 => 0.5357 moles

hope this helps!

5 0

4 years ago

Balance the following reaction in KOH (under basic conditions). What are the coefficients in for C3H8O2 and KMnO4 in the balance

GrogVix [38]

Answer:

Coefficient of $C_3H_8O_2=3$

Coefficient of $KMnO_4$ =8

Explanation:

We are given that a reaction in which $C_3H_8O_2$ reacts with $KMnO_4$

We have to find the coefficient of each reactants in balanced reaction

$3C_3H_8O_2(aq)+8KMnO_4(aq)\rightarrow 3C_3H_2O_4K_2(aq)+8MnO_2(aq)+2KOH+8H_2O$

Coefficient is defined the constant value multiplied with a reactant in a reaction.

Coefficient of $C_3H_8O_2$ =3

Coefficient of $KMnO_4=8$

Coefficient of $C_3H_2O_4K_2=3$

Coefficient of $MnO_2=8$

Coefficient of $H_2O=8$

Coefficient of KOH=2

Hence, Coefficient of $C_3H_8O_2=3$ and coefficient of $KMnO_4=8$

7 0

3 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

8 0

4 years ago

Science help me pleasee :)

vladimir2022 [97]

Explanation:

during this reaction this will produce 2 molecules of No

8 0

3 years ago