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Alex
3 years ago
6

What defines a force divide by area

Chemistry
1 answer:
pychu [463]3 years ago
8 0
Pressure is defined as the force divided by the area perpendicular to the force over which the force is applied, or. A given force can have a significantly different effect depending on the area over which the force is exerted.
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Which chemical formula represents both an element and a molecule?
victus00 [196]

Answer:

Explanation:

element - a basic substance that can't be simplified (hydrogen, oxygen, gold, etc...) molecule - two or more atoms that are chemically joined together (H2, O2, H2O, C6H12O6, etc...) compound - a substance that contains more than one element (H2O, C6H12O6, etc...)

5 0
2 years ago
What solute particles are present in an aqueous solution of CH3COCH3?
Anuta_ua [19.1K]
PE, GO, XY - I am  probably wrong xoxoxoxoxxo
4 0
3 years ago
If 1.0 gram of hydrogen reacts with 19.0 grams of fluorine, then what is the percent by mass of fluorine in the compound that is
astraxan [27]
The reaction between hydrogen (H2) and fluorine (F2) is given below,
                                   H2 + F2 ---> 2HF
One mole of both hydrogen and fluorine yields to 2 moles of hydrogen fluoride. This can also be expressed as, 2 grams of hydrogen and 38 grams of fluorine will form 40 grams of hydrogen fluoride. From the given, only 20 grams of HF is formed with 19 g of it being fluorine. Thus, the percentage fluorine of the compound formed is 95%. 
8 0
3 years ago
Read 2 more answers
Consider 2H2 + O2 → 2H2O. To produce 1.2 g water, how many grams of H2 are required? Report to the correct number of significant
Elden [556K]

Answer:

0.133 mol (corrected to 3 sig.fig)

Explanation:

Take the atomic mass of H=1.0, and O=16.0,

no. of moles = mass / molar mass

so no. of moles of H2O produced = 1.2 / (1.0x2+16.0)

= 0.0666666 mol

From the equation, the mole ratio of H2:H2O = 2:2 = 1:1,

meaning every 1 mole of H2 reacted gives out 1 mole of water.

So, the no, of moles of H2 required should equal to the no, of moles of H2O produced, which is also  0.0666666 moles.

mass = no. of moles x molar mass

hence,

mass of H2 required = 0.066666666 x (1.0x2)

= 0.133 mol (corrected to 3 sig.fig)

3 0
3 years ago
(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct
Zarrin [17]

Explanation:

The given data is as follows.

   \Lambda^{o}_{m}(NaCl) = 1.264 \times 10^{-2}

   \Lambda^{o}_{m}(H-O=C-ONO) = 1.046 \times 10^{-2}

   \Lambda^{o}_{m}(HCl) = 4.261 \times 10^{-2}

Conductivity of monobasic acid is 5.07 \times 10^{-2} S m^{-1}

     Concentration = 0.01 mol/dm^{3}

Therefore, molar conductivity (\Lambda_{m}) of monobasic acid is calculated as follows.

                 \Lambda_{m} = \frac{conductivity}{concentration}

                                  = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}

                                 = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}

                                 = 5.07 \times 10^{-3} S m^{2} mol^{-1}

Also, \Lambda^{o}_{m} = \Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}

                            = 4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}

                            = 4.043 \times 10^{-2} S m^{2} mol^{-1}

Relation between degree of dissociation and molar conductivity is as follows.

               \alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}

                             = \frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}

                             = 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

Putting the values into the above formula we get the following.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

                        = \frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}

                        = 0.017973 \times 10^{-2}

                       = 1.7973 \times 10^{-4}

Hence, the acid dissociation constant is 1.7973 \times 10^{-4}.

Also, relation between pK_{a} and K_{a} is as follows.

                 pK_{a} = -log K_{a}

                              = -log (1.7973 \times 10^{-4})

                              = 3.7454

Therefore, value of pK_{a} is 3.7454.

                             

3 0
3 years ago
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