All categories

2 years ago

What is the molality of a solution that contains 109 grams of NaOH dissolved in 4.55kg of water? The molar mass of NaOH is 40.00

1 answer:

8 0

$C_{m}=\frac{n}{V}=\frac{m_{s}}{MV}\\\\
d=\frac{m}{V} \ \ \ \Rightarrow \ \ \ V=\frac{m}{d}=\frac{4550g}{1\frac{g}{cm^{3}}}=4550cm^{3}=4,55L\\\\
C_{m}=\frac{109g}{40\frac{g}{mol}*4,55L}\approx0,6M$

You might be interested in

3C2H402 atoms in formula

Vanyuwa [196]

3( 2 carbons + 4 hydrogen's + 2 oxygens)

3*(8) = 24

4 0

3 years ago

A histidine is involved in an interaction with a glutamic acid that stabilizes the charged form of the histidine, such that the

Greeley [361]

Answer:

pKa of the histidine = 9.67

Explanation:

The relation between standard Gibbs energy and equilibrium constant is shown below as:

$\Delta{G^0} =-RT \ln \frac{[His]}{[His+]}$

R is Gas constant having value = 0.008314 kJ / K mol

Given temperature, T = 293 K

Given, $\Delta{G^0}=15\ kJ/mol$

So, Applying in the equation as:-

$15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}$

Thus,

$15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}$

$\frac{[His]}{[His+]}=e^{\frac{15}{-0.008314\times 293}$

$\frac{[His]}{[His+]}=0.00211$

Also, considering:-

$pH=pKa+log\frac{[His]}{[His+]}$

Given that:- pH = 7.0

So, $7.0=pKa+log0.00211$

<u>pKa of the histidine = 9.67</u>

8 0

2 years ago

Which of the following is not a base

SVEN [57.7K]

The answer is H₃C₆H₅O₇ (OPTION D).

In essence, the question is asking which of the options is an acid or a neutral compound; thus not a base. A base is a compound that accepts free hydrogen ions while in aqueous solution and as such they usually lack hydrogen ions that can be released in solution. If you ionize all the compounds/molecules above, the only on that has free hydrogen ions is OPTION D (which is an organic acid and not a base).

3 0

3 years ago

¿Cuál es el % m/m de una disolución en que hay disueltos 22 g de soluto en 44 g de disolvente?

Archy [21]

The question is as follows: What is the% m / m of a solution in which 22 g of solute are dissolved in 44 g of solvent?

Answer: The% m/m of a solution in which 22 g of solute are dissolved in 44 g of solvent is 50%.

Explanation:

Given: Mass of solute = 22 g

Mass of solvent = 44 g

The percentage m/m is calculated using the following formula.

$Mass percentage = \frac{mass of solute}{mass of solvent} \times 100$

Substitute the values into above formula as follows.

$Mass percentage = \frac{mass of solute}{mass of solvent} \times 100\\= \frac{22 g}{44 g} \times 100\\= 50 percent$

Thus, we can conclude that the% m/m of a solution in which 22 g of solute are dissolved in 44 g of solvent is 50%.

5 0

2 years ago

16. A sample of nitrogen gas,

Lana71 [14]

find mol of N2 present using gas law equation

PV = nRT

P = pressure = 688/760 = 0.905 atm.

V = 100mL = 0.1L

n = ???

R = 0.082057

T = 565+273 = 838

Substitute:

0.905*0.1 = n*0.082057*838

n = 0.0905 / 68.76

n = 0.00132 mol N2

Molar mass N2 = 28 g/mol

0.00132 mol = 0.00132*28 = 0.037g N2 gas

4 0

3 years ago