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aleksandrvk [35]
2 years ago
9

Suppose you have 200.0 mL of a 0.750 M sodium hydroxide solution. How many

Chemistry
1 answer:
Nikolay [14]2 years ago
5 0

Answer:

0.15 moles of sodium hydroxide are in the solution.

Explanation:

Molarity is a unit for expressing concentration of solutions. Molarity is defined as the number of moles of solute per liter of solution.

The molarity of a solution is calculated by dividing the moles of the solute by the liters of the solution:

Molarity (M)=\frac{number of moles of solute}{volume}

Molarity is expressed in units (\frac{moles}{liter}).

So, a molarity of 0.750 M indicates that 0.750 moles are present in 1 L of solution. Then the following rule of three can be applied: if in 1000 mL (being 1 L = 1000 mL) there are 0.750 moles, in 200 mL how many moles are there?

amount of moles=\frac{200 mL*0.750 moles}{1000 mL}

amount of moles=0.15

<u><em> 0.15 moles of sodium hydroxide are in the solution.</em></u>

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Which compound would be expected to show intense IR absorption at 2710 and 1705 cm-1? (Ph = phenyl group)
Serhud [2]

Answer:

B. PhCHO

Explanation:

Every organic group shows a characteristic IR absorption at certain wavelength . With the help of these absorption spectra we can identify the group present on organic molecules .

The wave number of 2710 cm⁻¹ is absorbed by aldehyde bond stretching .

The wave number of 1705 cm⁻¹ is shown by conjugated aldehyde . So the most likely compound among given compounds is PhCHO .

7 0
2 years ago
The ph of a solution that contains 0.818 m acetic acid (ka = 1.76 x 10-5) and 0.172 m sodium acetate is __________.
crimeas [40]
PH is calculated using <span>Handerson- Hasselbalch equation,

                                      pH  =  pKa  +  log [conjugate base] / [acid]

Conjugate Base  =  Acetate (CH</span>₃COO⁻)
Acid                    =  Acetic acid (CH₃COOH)
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                                      pH  =  pKa  +  log [acetate] / [acetic acid]

We are having conc. of acid and acetate but missing with pKa,
pKa is calculated as,

                                     pKa  =  -log Ka
Putting value of Ka,
                                     pKa  =  -log 1.76 × 10⁻⁵

                                     pKa  =  4.75
Now,
Putting all values in eq. 1,
                     
                                     pH  =  4.75 + log [0.172] / [0.818]

                                     pH  =  4.072
4 0
2 years ago
According to reference table adv-10, which reaction will take place spontaneously?
olga_2 [115]
Missing table!! write the elements with the first letter of the symbol with Upper Caps letters!!!

http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm

<span>Ni2+ +Pb(s) → Ni(s) + Pb2+
</span>The potential of the oxidation of Pb(s) --> Pb2+(aq) is 0.126 V 
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<span>Add the two together and the potential for the reaction is -0.124 V (NO SPONTANEOUS THE SIGN IS NEGATIVE)

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5 0
2 years ago
A solution is prepared by adding 6.24 g of benzene (C 6H 6, 78.11 g/mol) to 80.74 g of cyclohexane (C 6H 12, 84.16 g/mol). Calcu
tekilochka [14]

Answer:

x_B=0.0769

m=0.990m

Explanation:

Hello,

In this case, we can compute the mole fraction of benzene by using the following formula:

x_B=\frac{n_B}{n_B+n_C}

Whereas n accounts for the moles of each substance, thus, we compute them by using molar mass of benzene and cyclohexane:

n_B=6.24g*\frac{1mol}{78.11g}=0.0799mol\\ \\n_C=80.74g*\frac{1mol}{84.16g} =0.959mol

Thus, we compute the mole fraction:

x_B=\frac{0.0799mol}{0.0799mol+0.959mol}\\ \\x_B=0.0769

Next, for the molality, we define it as:

m=\frac{n_B}{m_C}

Whereas we also use the moles of benzene but rather than the moles of cyclohexane, its mass in kilograms (0.08074 kg), thus, we obtain:

m=\frac{0.0799mol}{0.08074kg}=0.990mol/kg

Or just 0.990 m in molal units (mol/kg).

Best regards.

7 0
2 years ago
Which solution contains the largest number of moles of chloride ions?
elixir [45]
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Part (a): <span>30.00 ml of 0.100m Cacl2
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Based on the above calculations, the correct answer is (d)</span>
5 0
2 years ago
Read 2 more answers
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