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Vlada [557]
3 years ago
9

____ KClO3-----> ____ KCl + ____O3

Chemistry
2 answers:
djyliett [7]3 years ago
6 0

Answer:

explain what need to be done here pls? what do you need?

Explanation:

photoshop1234 [79]3 years ago
4 0
Explain? add more info
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A student needed to make a 3 g/L NaCl solution. The student weighed 3 g of NaCl in a beaker and measured 1 L of water in a 1L vo
hoa [83]

Answer:

The answer is in the explanation.

Explanation:

A solution is defined as the <em>homogeneous mixture </em>of a solute (In this case, NaCl) and the solvent (water).

To prepare 1L of the solution, the student can weigh the 3g of NaCl in the volumetric flask but need to add slowly water to dissolve the NaCl (That is very soluble in water). When all NaCl is dissolved the student must transfer the solution to the 1L volumetric flask. Then, you must add more water to the beaker until "Clean" all the solute of the beaker to transfer it completely to the volumetric flask.

8 0
2 years ago
Which phrase is the best definition of matter?
Verdich [7]

The amount of physical material in an object

4 0
3 years ago
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How did Ernest Rutherford’s experiment relate to J.J. Thomson’s work?
horsena [70]
Thomson realized that the accepted model of an atom did not account for negatively or positively charged particles. Therefore, he proposed a model of the atom which he likened to plum pudding. ... Rutherford with the assistance of Ernest Marsden and Hans Geiger performed a series of experiments using alpha particles.
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3 years ago
Find the volume of a rectangular prism that is<br> 25 cm by 10 cm by 15 cm.<br> height<br> length
Stolb23 [73]

Answer:

3750 cm.

Explanation:

You multiply the three side measurements to find the volume.

25cm·10cm·15cm

375cm·10cm

3750 cm.

<em><u>Hope this helps!</u></em>

4 0
3 years ago
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Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given
tekilochka [14]

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

K_a= 1.34*10^{-5

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝)  \alpha = \sqrt{\frac{K_a}{C} }

\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }

\alpha = \sqrt{4.9629*10^{-5}}

\alpha = 7.044*10^{-3

percentage% (∝) = 7.044*10^{-3}*100

= 0.704%

For Part B; where Concentration of B = 7.84*10^{-2 M

\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }

\alpha = \sqrt{1.709*10^{-4} }

\alpha = 0.0130\\

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= 1.92*10^{-2} M

\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }

\alpha = \sqrt{6.802*10^{-4}}

\alpha =0.02608

percentage% (∝) = 0.02608  × 100%

= 2.60%

7 0
3 years ago
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