How is a kilogram of cotton balls similar to a kilogram of bricks

Classify each of these reactions. A single reaction may fit more than one classification. Ba ( ClO 3 ) 2 ⟶ BaCl 2 + 3 O 2 Ba(ClO

Lena [83]

Answer: a) $Ba(ClO_3)_2\rightarrow BaCl_2+3O_2$ : Decomposition

b) $NaNO_2+HCl\rightarrow NaCl+HNO_2$ : double displacement

c) $CaO+CO_2\rightarrow CaCO_3$ : Synthesis (Combination)

d) $ZnSO_4+Mg\rightarrow MgSO_4+Zn$ : redox

Explanation:

Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

$Ba(ClO_3)_2\rightarrow BaCl_2+3O_2$

A double displacement reaction is one in which exchange of ions take place.

$NaNO_2+HCl\rightarrow NaCl+HNO_2$

Synthesis reaction is a chemical reaction in which two reactants are combining to form one product.

$CaO+CO_2\rightarrow CaCO_3$

Redox reaction is a type of chemical reaction in which oxidation and reduction takes place in one single reaction. The oxidation number of one element increases and the oxidation number of other element decreases.

$ZnSO_4+Mg\rightarrow MgSO_4+Zn$

4 0

2 years ago

4) The initial rate of the reaction between substances P and Q was measured in a series of

ASHA 777 [7]

Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

experiments and the following rate equation was deduced.

$rate = k[P]^{2} [Q]$

Complete the table of data below for the reaction between P and Q

Explanation:

Given rate of the reaction is:

$rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }$

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

$k=\frac{rate}{[P]^{2}[Q] } \\ =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4$

Second row:

2. Rate value:

$rate =0.4* (0.10)^{2} * (0.10)\\\\ =4.0*10^-3mol.dm^-3.s^-1$

3.Third row:

$[Q]=\frac{rate}{k.[P]^{2} } \\ =9.6*10^-3 / (0.4 *(0.40)^{2} \\ =0.15mol.dm^{-3}$

4. Fourth row:

$[P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}$

6 0

2 years ago

Thallium has two stable isotopes, 203tl and 205tl. knowing that the atomic weight of thallium is 204.4, which isotope is the mor

Elza [17]

You have to figure out a way to write the two unknown abundances in terms of one variable.

The total abundance is 1 (or 100%). So if you say the abundance for the first one is X then the abundance for the second one has to be 1-X (where X is the decimal of the percentage so say 0.8 for 80%).

203(X) + 205(1-X) = 204.4

Then you just solve for X to get the percentage for TI-203.
And then solve for 1-X to get the percentage for TI-205.

After that the higher percentage would be the most abundant.

203x + 205 - 205x = 204.4
-2x + 205 = 204.4
-2x = -0.6

x = 0.3
1-x = 0.7

Then the TI-205 would have the highest percentage and would be the most abundant.

7 0

2 years ago

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

musickatia [10]

Answer:

$T_2=335.42K=62.27^oC$