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Naily [24]
3 years ago
7

The formula for the table salt is NaCl. Is table salt ionic or covalent? Explain.

Chemistry
1 answer:
mel-nik [20]3 years ago
6 0

Answer:

ionic

it is a mixture of a nonmetal and a metal which is what makes ionic compounds

covalent is a nonmetal and nonmetal

Explanation:

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D. dishwashing Soap/Liquid
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True or false To obtain a salt, a metal must be combined with a hydroxide
Darya [45]

Answer:

False

Explanation:

The statement is implying that this is the only way to obtain a salt.  Any ionic compound is a salt.  For example, NaCl (table salt) is an ionic compound.  It is the combination of a nonmetal (Cl) and a metal (Na).

3 0
3 years ago
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How many grams of Cl are in 345 g of CaCl2
swat32
Hope this helps you.

3 0
3 years ago
A fixed quantity of gas at 21 °c exhibits a pressure of 752 torr and occupies a volume of 5.12 l. (a) calculate the volume the g
mariarad [96]
A. PV=constant
752 torr = 0.989 atm
(0.989 atm)(4.38 L) = (1.88 atm)V
V = 2.30 L

B. T/V = constant
(294 K)/(4.38 L) = (448 K)/V
V = 6.67 L


Hoped this helped ( I believe it's right, sorry if it's not )
7 0
3 years ago
Read 2 more answers
What is the heat of vaporization of ethanol, given that ethanol has a normal boiling point of 63.5°C and the vapor pressure of e
Serjik [45]

Explanation:

According to Clausius-Claperyon equation,

       ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}]

The given data is as follows.

         T_{1} = 63.5^{o}C = (63.5 + 273) K

                         = 336.6 K

        T_{2} = 78^{o}C = (78 + 273) K

                         = 351 K

         P_{1} = 1 atm,             P_{2} = ?

Putting the given values into the above equation as follows.    

        ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}]

       ln (\frac{1.75 atm}{1 atm}) = \frac{-\text{heat of vaporization}}{8.314 J/mol K} \times [\frac{1}{351 K} - \frac{1}{336.6 K}]

                      \Delta H = \frac{0.559}{1.466 \times 10^{-4}} J/mol

                                  = 0.38131 \times 10^{4} J/mol

                                  = 3813.1 J/mol

Thus, we can conclude that the heat of vaporization of ethanol is 3813.1 J/mol.

4 0
3 years ago
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