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3 years ago

What most likely to the inclusion of these elements in the periodic table

Chemistry

1 answer:

antiseptic1488 [7]3 years ago

3 0

New technology showed how to identify one of the elements

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calculate the water potential of a solution of 0.15m sucrose. the solution is at standard temperature.

Mrac [35]

Answer:

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

Explanation:

Water potential = Pressure potential + solute potential

$P_w=P_p+P_s$

$P_w=P_p+(-iCRT)$

We have :

C = 0.15 M, T = 273.15 K

i = 1

The water potential of a solution of 0.15 m sucrose= $P_w$

$P_p=0 bar$ (At standard temperature)

$P_s=-iCRT=-\times 1\times 8.314\times 10^{-2}bar L/mol K\times 273.15 K=-3.406 bar$

$P_w=0 bar+(-3.406 ) bar$

The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.

7 0

3 years ago

Which method would be best to separate a mixture of sand and gravel?

solniwko [45]

Here, as they both are physical things without any magnetic property & density difference, screening would be best.

In short, Your Answer would be Option C

Hope this helps!

7 0

3 years ago

Read 2 more answers

If the theoretical yield of a reaction is 0.110 g and the actual yield is 0.104 g , what is the percent yield?

Westkost [7]

Theoretical Yield is an Ideal yield with 100 % conversion of reactant to product. It is in fact a paper work.

While,

Actual Yield is the yield which is obtained experimentally. It is always less than theoretical yield because it is not possible to have 100% conversion of reactants into products. Even some amount of product is lost while handling it during the process.

Percentage Yield is Calculated as,

%age Yield = Actual Yield / Theoretical Yield × 100

Data Given:
Actual Yield = 0.104 g

Theoretical Yield = 0.110 g

Putting Values,

%age Yield = 0.104 g / 0.110 g × 100

%age Yield = 94.54 %

3 0

3 years ago

Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib

Dvinal [7]

Answer: The value of $K_c$ for the final reaction is $7.16\times 10^{25}$

Explanation:

The given chemical equations follows:

Equation 1: $H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1$

Equation 2: $HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2$

The net equation follows:

$S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c$

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=\frac{1}{K_1}\times \frac{1}{K_2}$