3 years ago

Thank you in advance!

Chemistry

1 answer:

Contact [7]3 years ago

7 0

Answer:

nobody can answer this question because you have to draw model and you can't do it on here sorry .

Explanation:

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What is the density of a substance with a mass of 0.0450 and a volume of 26.4?

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Answer:

$density \: = \frac{mass}{volume} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{0.0450}{26.4} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 0.0017 \\$

Explanation:

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4 years ago

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asambeis [7]

Answer:

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3 years ago

in the reaction C5H8 ____02--8H20, what coefficient should be placed in frount of 02 to balance the reaction?

Gekata [30.6K]

5CO2 should be in the blank spot.

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3 years ago

Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t

Rasek [7]

Answer:

$4.6*10^{-14}$ M

Explanation:

Concentration of $Cu^{2+} = [Cu(NO_3)_2]$ = 0.020 M

Constructing an ICE table;we have:

$Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}$

Initial (M) 0.020 0.40 0

Change (M) - x - 4 x x

Equilibrium (M) 0.020 -x 0.40 - 4 x x

Given that: $K_f =1.7*10^{13}$

$K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}$

$1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}$

Since x is so small; 0.40 -4x = 0.40

Then:

$1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}$

$1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}$

$1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x$

$8.704*10^9-4.352*10^{11}x =x$

$8.704*10^9 = 4.352*10^{11}x$

$x = \frac{8.704*10^9}{4.352*10^{11}}$

$x = 0.0199999999999540$

$Cu^{2+}= 0.020 - 0.019999999999954$

$Cu^{2+} = 4.6*10^{-14}$ M

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3 years ago

How many moles of MgS2O3 are in 223 g of the compound

Alexxandr [17]

Moles of MgS2O3 = 223/molar mass of MgS2O3

= 223/136.42
= 1.634 moles.

Hope this helps!

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3 years ago

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