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GalinKa [24]
4 years ago
12

Cabbage juice turns pink in solutions with pH below 7 and green in solutions with pH above 7. You drop some cabbage juice in a s

ink full of soapy dishes and in a glass of soda. The cabbage juice is green in the soapy water and pink in the soda. Which substance is basic?
Chemistry
1 answer:
Ann [662]4 years ago
8 0
<span>Basic substances, when dissolved in water, give pH above 7</span>
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How would a solution that is labeled 5.0 M would be read as?
cestrela7 [59]
The M stands for molar, so it would be 5.0 molar. is that what you need?
7 0
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What is the speed of an electromagnetic wave that has a frequency of 7.8x106<br> Hz?
Vinil7 [7]

Answer:

3. x 10⁸m/s

Explanation:

Given parameters:

Frequency of wave  = 7.8 x 10⁶Hz

Unknown:

Speed of the wave = ?

Solution:

All electromagnetic radiation has  a speed of 3. x 10⁸m/s in free space.

No matter the frequency and wavelength, they all have the same speed which is the same as the speed of light.

5 0
3 years ago
An aqueous solution is 0.387 m in hcl. what is the molality of the solution if the density is 1.23 g/ml?
vazorg [7]
<span>Assume you have 1.00 L (1000 mL) of solution. d = m / V m = d x V = 1.23 g/mL x 1000 mL = 1230 g of solution 0.387 mol/L x 1 L = 0.387 mol HCl 0.387 mol HCl x (36.5 g / 1 mol) = 14.1 g HCl mass of water = 1230 g solution - 14.1 g HCl = 1216 g H2O = 1.216 kg H2O molality = mol HCl / kg water = 0.387 mol / 1.216 kg = 0.318 mol/kg (or 0.318 molal)</span>
8 0
3 years ago
True or false according to the principle of base pairing hydrogen bonds could form only between adenine and cytosine
prohojiy [21]
Answer is: false.
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8 0
3 years ago
When solid calcium carbonate is heated, it decomposes according to the reaction: CaCO3(s) ⇄ CaO(s) + CO2(g) Kp = 0.50 A sample o
blagie [28]

Explanation:

Relation between K_{p} and K_{c} is as follows.

            K_{p} = K_{c} [RT]^{\Delta n}

Given,   temperature = 830^{o}C = (830 + 273) K = 1103 K

            R = 8.314 J/mol K

       \Delta n = 1 - 0 = 1

Now, putting the given values into the above formula as follows.

              K_{p} = K_{c} [RT]^{\Delta n}

                 0.5 = K_{c} \times (8.314 \times 1103)^{1}

                     K_{c} = 5.452 \times 10^{-3}

ICE table for the given reaction will be as follows.

                  CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)

Initial:           c                        -          -  

Equilibrium:  (c - x)               x         x

       K_{c} = [CO_{2}]

          5.452 \times 10^{-3} = x

Hence, same amount of CaO is produced.

Moles of CaO = 5.452 \times 10^{-3}

    Mass of CaO = 5.452 \times 10^{-3} \times 57 g/mol

                          = 0.310 g

Thus, we can conclude that 0.310 g of CaO produced when equilibrium is established.

8 0
3 years ago
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