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2 years ago

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Test questions Accidently did this

2 answers:

PSYCHO15rus [73]2 years ago

6 0

I don’t see any questions

USPshnik [31]2 years ago

5 0

Answer:

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Explanation:

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An ice cube has a mass of 25 grams. The ice cube is placed in a glass and melts. What is TRUE about the ice cube? A) The melted

masya89 [10]

The answer is C. Hope this helped

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3 years ago

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

3 years ago

A rigid vessel contains water at 0.15 MPa (at 25 °C). The vessel is heated to the critical point of water. Calculate the fractio

pishuonlain [190]

Answer:

0.8078 Kg

Explanation:

Pressure of water = 0.15 MPa = 1.5 bar

At critical point of water ,temperature = 647 K=374°C

From the ideal gas equation

P×V= m×R×T

Let us assume volume = 1 m^3

1.5 x 105 x 1 = m x 287 x 647

m= 0.8078 kg

the fraction of mass of liquid at 25°C.

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3 years ago

(2. Which of the following best describes humus?

Harlamova29_29 [7]

Answer:

A. Decomposed organic matter

Explanation:

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2 years ago

Which Two of the following are pure substances*

Snowcat [4.5K]

Answer:

I think A and B

Explanation:

5 0

3 years ago

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