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Katarina [22]
3 years ago
6

Be sure to answer all parts. identify and label the species in each reaction. (a) nh4+(aq) + h2o(l) ⇌ nh3(aq) + h3o+(aq) acid ba

se acid base conjugate acid conjugate base conjugate acid conjugate base (b) cn−(aq) + h2o(l) ⇌ hcn(aq) + oh−(aq) acid base acid base conjugate acid conjugate base conjugate acid conjugate base
Chemistry
1 answer:
quester [9]3 years ago
5 0
A)

NH⁴⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃0⁺<span>(aq)

- acid </span>a species that able to donate (H+): NH⁴⁺
- base a species that is able to accept a proton (H+): H₂O
- conjugate base a species formed when acid donates a proton (H+): NH₃
- conjugate acid a species formed by a base accepts a proton (H+): H₃0⁺

b)

CN⁻(aq) + H₂O(l) ⇌ HCN(aq) + OH⁻(aq)

- base a species that is able to accept a proton (H+): CN⁻
- acid a species that able to donate (H+): H₂O
- conjugate acid a species formed by a base accepts a proton (H+): HCN
- conjugate base a species formed when acid donates a proton (H+): OH⁻

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

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Width of the base  =  w

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We have:

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

Square based pyramid:

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

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

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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

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

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3 years ago
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