When a substance was dissolved in water, the temperature of the water increased. This process is described as

Give the formula of the conjugate base of HSO4–

mr Goodwill [35]

The answer is So^2-4

7 0

3 years ago

H3C6H507(aq) + 3NaHCO3(aq) - Na3C6H5O7(aq) + 3C02(g) + 3__

zavuch27 [327]

Answer: 3<u> H₂O</u>

<u />

Explanation:

This is an example of a common chemical reaction

ACID + CARBONATE --> SALT + WATER + CARBON DIOXIDE which implires

CITRIC ACID + SODIUM BICARBONATE ----> SODIUM CITRATE + WATER + CARBON DIOXIDE .

The full chemical equation is given by

H3C6H5O7(aq) + 3NaHCO3(aq) --> Na3C6H5O7(aq) + 3CO2(g) + 3<u>H2O</u>(l)

The reaction produces a fizz due to the release of carbon dioxide which can be also seem when an Alka-Seltzer® tablet is dissolved in water.

6 0

4 years ago

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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

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

$=13.20 cm/s$

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

3 0

3 years ago

The pressure of a sample of argon gas was increased from 3.14 atm to 7.98 at a constant temperature. If the final volume of argo

noname [10]

Answer:

Explanation:

The initial volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume.

Since we're finding the initial volume

$V_1 = \frac{P_2V_2}{P_1} \\$

We have

$V_1 = \frac{7.98 \times 14.2}{3.14} = \frac{113.316}{3.14} \\ = 36.0878...$

We have the final answer as

Hope this helps you

8 0

3 years ago

Energy release is to condensation as energy input is to _________.

melamori03 [73]

Answer:

sublimation

Explanation:

the other choices release energy

7 0

3 years ago

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