HELPPP

an aqueous solution contains 1.2mM of total ions. if the solution is NaCl (aq), what is the concentration of chloride ions?

Genrish500 [490]

Explanation:

As the total concentration is given as 1.2 mM. And, it is also given that salt present in the solution is NaCl.

As sodium chloride is an ionic compound so, when it is added to water then it will dissociate into sodium and chlorine ions as follows.

$NaCl \rightarrow Na^{+} + Cl^{-}$

So, it means in total there will be formation of 2 ions when one molecules of NaCl dissociates.

Therefore, concentration of chlorine ions will be calculated as follows.

Concentration of $Cl^{-}$ ions = $\frac{1.2mM}{2}$

= 0.6 mM

Thus, we can conclude that the concentration of chloride ions is 0.6 mM.

5 0

3 years ago

55 L of a gas at 25oC has its temperature increased to 35oC. What is its new volume?

ladessa [460]

Answer:

Approximately 56.8 liters.

Assumption: this gas is an ideal gas, and this change in temperature is an isobaric process.

Explanation:

Assume that the gas here acts like an ideal gas. Assume that this process is isobaric (in other words, pressure on the gas stays the same.) By Charles's Law, the volume of an ideal gas is proportional to its absolute temperature when its pressure is constant. In other words

$\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1}$ ,

where

$V_2$ is the final volume,
$V_1$ is the initial volume,
$T_2$ is the final temperature in degrees Kelvins.
$T_1$ is the initial temperature in degrees Kelvins.

Convert the temperatures to degrees Kelvins:

$T_1 = \rm 25^{\circ}C = (25 + 273.15)\; K = 298.15\; K$ .

$T_2 = \rm 35^{\circ}C = (35 + 273.15)\; K = 308.15\; K$ .

Apply Charles's Law to find the new volume of this gas:

$\displaystyle V_2 = V_1\cdot \frac{T_2}{T_1} = \rm 55\;L \times \frac{308.15\; K}{298.15\; K} = 56.8\; L$ .

8 0

3 years ago

A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of

Stella [2.4K]

Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

$n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC$

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

$n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH$

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

$m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO$

4. Compute the moles of oxygen by using its molar mass:

$n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO$

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

$C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1$

6. Search for the closest whole number (in this case multiply by 2):

$C_3H_6O_2$

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

$M=12*3+1*6+16*2=74g/mol$

Which is about three times in the molecular formula, for that reason, the actual formula is:

$C_9H_{18}O_6$

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

6 0

3 years ago

How many moles are needed

geniusboy [140]

Answer: amount of substance n = cV = 2.0 mol/l x 0.25 l = 0.50 mol

Explanation:

7 0

2 years ago

Atom is anything that has mass and occupies space. True or false

rosijanka [135]

Answer:

False.

Explanation:

Matter is anything that has mass and occupies space, not atoms.

4 0

3 years ago