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Colt1911 [192]
3 years ago
15

Can you unhelpful people answer this question already i'll mark you the brainlest. I'll mark you the brainlest if you answer the

question.What is the independent variable shown on this graph
A. only volume (cm³) is the independent variable

B. both volume (cm³) and mass (grams) are independent variable

C. not enough information

D. the slope of the trend line

E. only mass (grams) is the independent variable

Chemistry
1 answer:
Genrish500 [490]3 years ago
5 0
A. only volume (cm^3)

Explanation:
The independent variable is typically represented as x. It is the variable that you change during the experiment. In this graph, as volume increases, it increases the mass too. Hope this helps!
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Assuming that an acetic acid solution is 12% by mass and that the density of the solution is 1.00 g/mL, what volume of 1 M NaOH
Doss [256]

Explanation:

Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.

Now, calculate the number of moles of acetic acid as its molar mass is 60 g/mol.

    No. of moles = \frac{mass}{\text{molar mass}}

                           = \frac{12 g}{60 g/mol}

                           = 0.2 mol

Molarity of acetic acid is calculated as follows.

              Density = \frac{mass}{volume}

                 1 g/ml = \frac{100 g}{volume}

                    volume = 100 ml

Hence, molarity = \frac{\text{no. of moles}}{volume}

                           = \frac{0.2 mol}{0.1 L}

                           = 2 mol/l

As reaction equation for the given reaction is as follows.

     NaOH + CH_{3}COOH \rightarrow CH_{3}COONa + H_{2}O

So,          moles of NaOH = moles of acetic acid

Let us suppose that moles of NaOH are "x".

          x \times 1 M = 10 mL \times 2 M     (as 1 L = 1000 ml)

                        x = 20 L

Thus, we can conclude that volume of NaOH required is 20 ml.                    

6 0
4 years ago
In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe
const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = 10^{-9} m.

Hence, 0.283 nm = 0.283 \times 10^{-9} m

  • Formula for coulombic energy is as follows.

             U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}

where,   e = 1.6 \times 10^{-19} C

            \epsilon_{o} = 8.85 \times 10^{-12}

          U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}

                         = 1.423 \times 10^{-18} J

  • As 1 eV = 1.6 \times 10^{-19} J

So,       1 J = \frac{1 eV}{1.6 \times 10^{-19}}

Hence,    U = \frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}

                   = 8.9 eV

  • Also,   1 J = \frac{10^{-3} kJ}{6.022 \times 10^{23}mol}

                = 1.67 \times 10^{-27} kJ/mol

Therefore, U = 1.423 \times 10^{-18} J \times 1.67 \times 10^{-27} kJ/mol

                     = 2.37 \times 10^{-45} kJ/mol

7 0
3 years ago
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