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sweet-ann [11.9K]
3 years ago
9

Complete the paragraph below. Do it using your Science journal or notebook

Chemistry
1 answer:
svetlana [45]3 years ago
3 0

Answer:

SUSPENSION IS A KIND OF MIXTURE where particlea are visible to the naked eyes AS IT settled AT THE BOTTOM WHEN LEFT UNDISTURBED.

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The reaction of Pb+4 and O-2 produces the compound of
vovikov84 [41]

Answer:

Pb₂O₄

                               

Explanation:

The given species are:

        Pb⁴⁺               O²⁻  

Now, to solve this problem, we use the combining powers which corresponds to the number of electrons usually lost or gained or shared by atoms during the course of a chemical combination.

                                          Pb⁴⁺               O²⁻  

Combining power               4                    2

Exchange of valencies        2                    4

Now the molecular formula is  Pb₂O₄

                               

4 0
2 years ago
Convert 512 kilograms to milligrams.<br><br> 0.000512<br> 0.512<br> 512,000<br> 512,000,000
olga55 [171]
The answer is D! 512000000
5 0
3 years ago
Read 2 more answers
What else is produced during the combustion of butane, C4H10?<br><br> 2C4H10 + 13O2 → <br> + 10H2O
Neko [114]
<h3>Answer:</h3>

8CO₂

<h3>Explanation:</h3>

We are given;

  • Butane, C₄H₁₀
  • Butane is a hydrocarbon in the homologous series known as alkane.

We are required to determine the other product produced in the combustion of butane apart from water.

  • We know that the complete combustion of alkane yields carbon dioxide and water.
  • Therefore, combustion of butane will yield carbon dioxide and water.
  • The balanced equation for the complete combustion of butane will be;

       2C₄H₁₀ + 13O₂ →  8CO₂ + 10H₂O

8 0
3 years ago
Read 2 more answers
A)<br>Name the following compounds<br>CH, -CH-CH<br>(i)<br>CH, Br<br>-​
liq [111]

Answer:

methyl, ethyl,

Explanation:

that should be the case

8 0
2 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔
Vilka [71]

Answer : The correct option is, (C) 1.1

Solution :  Given,

Initial moles of N_2O_4 = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

The given equilibrium reaction is,

                           N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initially                      c                 0

At equilibrium   (c-c\alpha)           2c\alpha

The expression of K_c will be,

K_c=\frac{[NO_2]^2}{[N_2O_4]}

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

where,

\alpha = degree of dissociation = 40 % = 0.4

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

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}

K_c=1.066\aprrox 1.1

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0
2 years ago
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