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4 years ago

Fill in the [?]: 12,341 us = [?] S

Chemistry

1 answer:

BigorU [14]4 years ago

3 0

Currency, is my best assumption without a subject to determine.

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Balance the following reaction: CH4 + O2 > CO2 + H20 Enter your answer in the space provided.

KATRIN_1 [288]

Ignore the delta H. It's thermodynamics related problem I answered a few minutes ago

6 0

4 years ago

Read 2 more answers

Which students propose a conservation strategy that represents reducing the use of

andre [41]

Answer:

student A

Explanation:

5 0

3 years ago

Calculate the volume in milliliters of a 0.420mol / L barlum chlorate solution that contains 25.0 g of barium chlorate (Ba(ClO 3

Pachacha [2.7K]

<u>Answer:</u> The volume of barium chlorate is 195.65 mL

<u>Explanation:</u>

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of barium chlorate = 25.0 g

Molar mass of barium chlorate = 304.23 g/mol

Molarity of solution = 0.420 mol/L

Volume of solution = ?

Putting values in above equation, we get:

$0.420mol/L=\frac{25.0\times 1000}{304.23\times V}\\\\V=\frac{25.0\times 1000}{304.23\times 0.420}=195.65mL$

Hence, the volume of barium chlorate is 195.65 mL

6 0

3 years ago

It is difficult to break the ionic bonds in a compound because of the...

mestny [16]

The answer to the first one is C. Strong attraction between ions

The answer to the second one is A. High melting points

Hope this helps :)

6 0

4 years ago

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0.0200 M Fe3+ is initially mixed with 1.00 M oxalate ion, C2O42-, and they react according to the equation: Fe3+(aq) + 3 C2O42-(

Studentka2010 [4]

Answer : The concentration of $Fe^{3+}$ at equilibrium is 0 M.

Solution : Given,

Concentration of $Fe^{3+}$ = 0.0200 M

Concentration of $C_2O_4^{2-}$ = 1.00 M

The given equilibrium reaction is,

$Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)$

Initially conc. 0.02 1.00 0

At eqm. (0.02-x) (1.00-3x) x

The expression of $K_c$ will be,

$K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}$

$1.67\times 10^{20}=\frac{(x)^2}{(1.00-3x)^3\times (0.02-x)}$

By solving the term, we get:

$x=0.02M$

Concentration of $Fe^{3+}$ at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M

Therefore, the concentration of $Fe^{3+}$ at equilibrium is 0 M.

4 0

3 years ago