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2 years ago

Can someone help me with questions 8 and 9 plz. I will really appreciated it

Chemistry

1 answer:

puteri [66]2 years ago

5 0

Answer:

8. A boat, cars, airplanes

9. It's safer to practice techniques on life sized models of human organs than practicing on an actual human. If you mess up on a model it doesn't matter, messing up on a human can put their life at risk.

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What is the weight of an object that has a mass of 100 kg? • 100 N • 200 N • 980 N

Cloud [144]

Answer:i really know it is 29.09

Explanation:

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2 years ago

For each of the salts below, match the salts that can be compared directly, using Ksp values, to estimate solubilities.

Anni [7]

Answer:

1 - Salts required that can be compared directly, using Ksp values, to estimate solubilities for copper (II) sulfide are - $CaSO_3$ (Option b) , $BaCrO_4$ (Option c) and $CaS$ (Option d)

2 - Salts that can be compared directly, using Ksp values, to estimate solubilities for zinc hydroxide are - $Mg(OH)_2$ (Option a)

Explanation:

$K_s_p$ values can be used to compare the solubility of salts that produce ions in a 1:1 ratio.

$CuS$ {copper (II)sulfide} dissociates with ions in ratio 1:1 .

Because the salts $CaSO_3$ , $BaCrO_4$ and $CaS$ have a 1:1 ion ratio, the solubilities of options b, c, and d can be compared to salt 1.

$K_s_p$ Values can be used to compare the solubility of salts that produce ions in a 1:2 or 2:1 ratio.

2. $Zn(OH)_2$ dissociates into ions in ratio 1:2 .

Because the ions in the salt $Mg(OH)_2$ are in a 1:2 ratio, the solubility of salt 2 can be compared to that of salt 'a'.

Hence , Salt 1 is matched with options b , c , d.

Salt 2 is matched with option a.

4 0

2 years ago

Both "Ode to the West wind" and "Ode for Melancholy"

Rama09 [41]

The answer that would best complete the given statement above would be option A. Both "Ode to the West wind" and "Ode for Melancholy" praise something non-human. Hope this answers your question. Have a great day ahead!

5 0

2 years ago

Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol

Sonbull [250]

Answer: The value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

Explanation:

We are given:

Initial moles of nitrogen gas = 1.30 moles

Initial moles of hydrogen gas = 1.65 moles

Equilibrium moles of ammonia = 0.100 moles

Volume of the container = 1.00 L

For the given chemical equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

Initial: 1.30 1.65

At eqllm: 1.30-x 1.65-3x 2x

Evaluating the value of 'x'

$\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}$

Equilibrium moles of nitrogen gas = $(1.30-x)=(1.30-0.05)=1.25mol$

Equilibrium moles of hydrogen gas = $(1.65-x)=(1.65-0.05)=1.60mol$

Putting values in above expression, we get:

$K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}$

Calculating the $K_c'$ for the given chemical equation:

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

$K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2$

Hence, the value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

8 0

2 years ago

Read 2 more answers

______ will have a higher entropy than 50.0 g CO2 (g) at 56 oC.

Tomtit [17]

1) The more mass is the more entropy , because there are more particles, there is disorder.
2) Than higher temperature --- the more entropy.
3) Gas has more disorder than liquid, so gas has more entropy.
So, correct answer is E.

5 0

2 years ago