The transfer of thermal energy for a warmer object too a cooler one. 10 points

Will mark Brainliest!! In a solution of salt and water, which component is the solute?

Alexxx [7]

Answer salt

Explanation:

5 0

4 years ago

What is the freezing point (in degrees Celcius) of 3.75 kg of water if it contains 189.9 g of C a B r 2?

blondinia [14]

Answer:

The freezing point of the solution is -1.4°C

Explanation:

Freezing point decreases by the addition of a solute to the original solvent, freezing point depression formula is:

ΔT = kf×m×i

Where Kf is freezing point depression constant of the solvent (1.86°C/m), m is molality of the solution (Moles CaBr₂ -solute- / kg water -solvent) and i is Van't Hoff factor.

Molality of the solution is:

-moles CaBr₂ (Molar mass:

189.9g ₓ (1mol / 199.89g) = 0.95 moles

Molality is:

0.95 moles CaBr₂ / 3.75kg water = 0.253m

Van't hoff factor represents how many moles of solute are produced after the dissolution of 1 mole of solid solute, for CaBr₂:

CaBr₂(s) → Ca²⁺ + 2Br⁻

3 moles of ions are formed from 1 mole of solid solute, Van't Hoff factor is 3.

Replacing:

ΔT = kf×m×i

ΔT = 1.86°C/m×0.253m×3

ΔT = 1.4°C

The freezing point of water decreases in 1.4°C. As freezing point of water is 0°C,

<h3>The freezing point of the solution is -1.4°C</h3>

4 0

3 years ago

Read 2 more answers

The Quiver tree grows in Southern Africa. Which of the following plant adaptations is likely to prevent these trees from dying o

Gekata [30.6K]

Answer:

The correct option is Ability to store water in leafy structures to prevent excess evaporation.

Explanation:

Desert plants need to have adaptations so that they can withstand the harsh conditions of the environment. In order to withstand the harsh temperatures of the desert, a quiver tree has leafy structures which help it to store water for long time. In this way, water is prevented to be lost from the tree due to evaporation. Hence, it is an effective adaptation to live in desert conditions.

The Quiver tree also has white powder on is branches to hide from the Sun's rays.

4 0

4 years ago

Please help I’m timed !!!

Rashid [163]

Answer:

b. $\% diss =4.4x10^{-2}\%$

Explanation:

Hello there!

In this case, given the ionization reaction of HClO as weak acid:

$HClO\rightleftharpoons H^++ClO^-$

We can write the equilibrium expression as shown below:

$Ka=3.5x10^{-8}=\frac{[H^+][ClO^-]}{[HClO]}$

In such a way, via the definition of x as the reaction extent, we can write:

$3.5x10^{-8}=\frac{x^2}{[HClO]}$

As long as Ka<<<<1 so that the x on the bottom can be neglected. Thus, we solve for x as shown below:

$x=\sqrt{3.5x10^{-8}*0.18} =\\\\x=7.94x10^{-5}M$

And finally the percent dissociation:

$\% diss=\frac{x}{[HClO]} *100\%\\\\\% diss=\frac{7.94x10^{-5}M}{0.18}*100\% \\\\\% diss =0.044\%=4.4x10^{-2}\%$

Which is choice b.

Best regards!

6 0

3 years ago

While ethanol (CH3CH2OH) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by

Anna [14]

Answer: The number of moles of ethanol after equilibrium is reached the second time is 11. moles.

Explanation:

We are given:

Initial moles of ethene = 34 moles

Initial moles of water vapor = 15 moles

The chemical equation for the formation of ethanol follows:

$CH_2=CH_2+H_2O\rightleftharpoons CH_3CH_2OH$

Initial: 34 15

At eqllm: 34-x 15-x x

We are given:

Equilibrium moles of ethene = 24 moles

Equilibrium moles of water vapor = 5 moles

Calculating for 'x'. we get:

$34-x=24\\\\x=10$

Volume of container = 100.0 L

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CH_3CH_2OH]}{[CH_2=CH_2][H_2O]}$ .......(1)

$[CH_3CH_2OH]=\frac{10}{100}=0.1M$

$[CH_2=CH_2]=\frac{24}{100}=0.24M$

$[H_2O]=\frac{5}{100}=0.05M$

Putting values in expression 1, we get:

$K_c=\frac{0.1}{0.24\times 0.05}\\\\K_c=8.3$

Now, 11 moles of ethene gas is again added and equilibrium is re-established, we get:

The chemical equation for the formation of ethanol follows:

$CH_2=CH_2+H_2O\rightleftharpoons CH_3CH_2OH$

Initial: 24+11 5 10

At eqllm: 35-x 5-x 10+x

$[CH_3CH_2OH]=\frac{10+x}{100}$

$[CH_2=CH_2]=\frac{35-x}{100}$

$[H_2O]=\frac{5-x}{100}$

Putting values of in expression 1, we get:

$8.3=\frac{\frac{(10+x)}{100}}{\frac{(35-x)}{100}\times \frac{(5-x)}{100}}\\\\8.3=\frac{(10+x)\times 100}{(35-x)\times (5-x)}\\\\x^2-52x+55=0\\\\x=50.9,1.1$

The value of 'x' cannot exceed '35', so the numerical value of x = 50.9 is neglected.

Moles of ethanol = $(10+x)=10+1.1=11.$

Hence, the number of moles of ethanol after equilibrium is reached the second time is 11. moles.

8 0

3 years ago