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otez555 [7]
3 years ago
5

Write the balanced chemical Equation for the following when they are heated ​

Chemistry
1 answer:
ale4655 [162]3 years ago
4 0

Answer:

i. Lead nitrate:

2Pb (NO3)2 Δ= 2PbO+4NO2+O2

ii. Potassium chlorate:

2KClO3 → 2KCl + 3O2↑

iii. clacium carbonate:

CaCO3 + 2HCl -> CaCl2 + CO2 + H2O

iv. cupric carbonate :

CuCO3 → CuO + CO2↑

Hope this helped

All the best!!

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3 years ago
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4. How many milligrams are in 5.25 x 10-13 kg?<br><br> the “-13” is an exponent
rusak2 [61]

5. 25 x 10⁻⁷mg

Explanation:

This is mass conversion from mg to kg;

The kg is a quantity of mass used to measure the amount of matter in a substance.

   Given mass = 5.25 x 10⁻¹³kg

The kilo-  is a prefix that denotes 10³

  therefore;

         1000g = 1kilogram

 the milli-  is a prefix that denotes 10⁻⁻³

       1000mg = 1g

Now that we know this, we can convert:

   5.25 x 10⁻¹³kg  x \frac{1000g}{1kg}  x \frac{1000mg}{1g}   =  5. 25 x 10⁻¹³ x 10⁶mg

      =  5. 25 x 10⁻⁷mg

learn more:

Conversion brainly.com/question/1548911

#learnwithBrainly

8 0
3 years ago
Build a prototype or test model of the invention​
KIM [24]
What invention I want to help you but I am confused
8 0
3 years ago
A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles
alexgriva [62]

Answer : The pH of the solution is, 3.41

Explanation :

First we have to calculate the moles of HF.

\text{Moles of HF}=\text{Concentration of HF}\times \text{Volume of solution}

\text{Moles of HF}=0.250M\times 1.50L=0.375mol

Now we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (6.8\times 10^{-4})

pK_a=4-\log (6.8)

pK_a=3.17

The reaction will be:

                             HF+OH^-\rightleftharpoons F^-+H_2O

Initial moles     0.375     0.100   0.375

At eqm.   (0.375-0.100)      0     (0.375+0.100)

                     = 0.275                    = 0.475

Now we have to calculate the pH of solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[F^-]}{[HF]}

Now put all the given values in this expression, we get:

pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}]

pH=3.41

Thus, the pH of the solution is, 3.41

8 0
3 years ago
Explain why a solution can have many solutes, but only one solvent.
kkurt [141]
A solution has to have at least one solute, but there can be any number more than that. <span>There is only one solvent in any solution, whatever liquid is present in largest amount.</span>
7 0
3 years ago
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