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3 years ago

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All of the following conditions of STP are true except A. 101.3 kPa B. 273.15 K C. 22.4 L D. 3.81 kPa

1 answer:

kaheart [24]3 years ago

8 0

3.81 kpa is the condition which is not true at STP

According to IUPAC the standard temperature and pressure that is STP the temperature is 273.15 k or 0 degrees celsius . and the absolute temperature of 101.325 Kpa or 1 atm. In addition at STP the volume of ideal gas is 22.4

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Working on-board a research vessel somewhere at sea, you have (carefully) isolated 12.5 micrograms (12.5 ×10–6 g) of what you ho

german

Answer:

The value is $Z = 311.33 \ g/mol$

Explanation:

From the question we are told that

The mass of saxitoxin is $m = 12.5 mg = 12.5 * 10^{-6} g$

The volume of water is $V = 3.10 mL = 3.10 *10^{-3} L$

The osmotic pressure is $P = 0.236 = \frac{0.236}{760} = 3.105 * 10^{-4} atm$

The temperature is $T = 19^oC = 19 + 273 = 292 \ K$

Generally the osmotic pressure is mathematically represented as

$P = C * T * R$

Here R is the gas constant with value

$R = 0.0821 ( L .atm /mol. K)$

and C is the concentration of saxitoxin

So

$3.105 * 10^{-4} = C * 0.0821 * 292$

$C = 1.295 *10^{-5} mol/L$

Generally the number of moles of saxitoxin is mathematically represented as

$n = C * V$

=> $n = 1.295 *10^{-5} *3.10 *10^{-3}$

=> $n = 4.015 *10^{-8} \ mol$

Generally the molar mass of saxitoxin is mathematically represented as

$Z = \frac{m}{n}$

=> $Z = \frac{12.5 * 10^{-6}}{ 4.015 *10^{-8}}$

=> $Z = 311.33 \ g/mol$

5 0

3 years ago

Potassium chloride is used as a substitute for sodium chloride for individuals with high blood pressure. Identify the limiting r

djyliett [7]

Answer:

* Limiting reagent: potassium.

* $m_{Cl_2}^{excess}=2.47gCl_2$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2K(s)+Cl_2(g)\rightarrow 2KCl$

In such a way, the reacting moles of chlorine are:

$n_{Cl_2}=7.00g*\frac{1molCl_2}{70.9gCl_2}=0.0987molCl_2$

Now, the amount of chlorine gas that would react with 5.00 g of potassium result:

$n_{Cl_2}^{consumed}=5.00g*\frac{1molK}{39.1gK}*\frac{1molCl_2}{2molK}=0.0639molCl_2$

Thus, the chlorine will be in excess and the potassium will be the limiting reagent. Therefore, the mass of excess chlorine turns out:

$m_{Cl_2}^{excess}=(0.0987-0.0639)molCl_2*\frac{70.9gCl_2}{1molCl_2}=2.47gCl_2$

Best regards.

4 0

3 years ago