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Natali5045456 [20]
3 years ago
9

What is the concentration of Agt in a 1.2 x 10-4 solution of Ag2CO3? (To write your answer using scientific notation use 1.0E-1

instead of 1.0 x 10-1)
Chemistry
1 answer:
MrRa [10]3 years ago
7 0

Answer:

2.4E-4

Explanation:

Hello,

By applying the following mass-mole relationship, the concentration could be computed as follows (assuming molarity as long as it isn't specified), since in the silver carbonate two silver molecules are present:

[Ag]=1.2x10^{-4}\frac{molAg_2CO_3}{L} *\frac{2mol Ag}{1 mol Ag_2CO_3}=2.4x10^{-4}\frac{mol Ag}{L}

Best regards.

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An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, h
vaieri [72.5K]

Answer:

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

Explanation:

Length of the base = l

Width of the base  =  w

Height of the pyramid = h

Volume of the pyramid = V=\frac{1}{3}lwh

We have:

Rate at which water is filled in cube = \frac{dV}{dt}= 45 cm^3/s

Square based pyramid:

l = 6 cm, w = 6 cm, h = 13 cm

Volume of the square based pyramid = V

V=\frac{1}{3}\times l^2\times h

\frac{l}{h}=\frac{6}{13}

l=\frac{6h}{13}

V=\frac{1}{3}\times (\frac{6h}{13})^2\times h

V=\frac{12}{169}h^3

Differentiating V with respect to dt:

\frac{dV}{dt}=\frac{d(\frac{12}{169}h^3)}{dt}

\frac{dV}{dt}=3\times \frac{12}{169}h^2\times \frac{dh}{dt}

45 cm^3/s=3\times \frac{12}{169}h^2\times \frac{dh}{dt}

\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times h^2}

Putting, h = 4 cm

\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times (4 cm)^2}

=13.20 cm/s

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