A + B > AB is what kind of reaction?

Stels [109]

Answer: The number to the left of AC should be 6.

Explanation: The balanced chemical reaction is one in which the number of atoms of each element on the reactant side must be equal to the number of atoms on product side.

The given equation $A_2B + DC_3 \rightarrow AC + D_2B_3$ is unbalanced as the atoms on the reactant side are not same as number of atoms on product side. This equation is called as skeletal equation.

The balanced chemical equation is :

$3A_2B + 2DC_3 \rightarrow 6AC + D_2B_3$

Thus the number to the left of AC is 6.

6 0

2 years ago

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A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4

Stella [2.4K]

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron = 560 J/(kg.K)

$c_1$ = specific heat of water = 4186 J/(kg.K)

$m_1$ = mass of iron = 825 g

$m_2$ = mass of water = 40 g

$T_f$ = final temperature of water and iron = ?

$T_1$ = initial temperature of iron = $352^oC=273+352=625K$

$T_2$ = initial temperature of water = $20^oC=273+20=293K$

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

$(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)$

$T_f=537.12K$

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

8 0

2 years ago

Complete the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (Ty

anastassius [24]

Answer :

(A) The dissociation reaction of $HC_2H_3O_2$ will be:

$HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

(B) The dissociation reaction of $Co(H_2O)_6^{3+}$ will be:

$Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}$

(C) The dissociation reaction of $CH_3NH_3^+$ will be:

$CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}$

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

(A) The dissociation reaction of $HC_2H_3O_2$ will be:

$HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)$

The equilibrium expression of $HC_2H_3O_2$ will be:

$K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

(B) The dissociation reaction of $Co(H_2O)_6^{3+}$ will be:

$Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)$

The equilibrium expression of $Co(H_2O)_6^{3+}$ will be:

$K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}$

(C) The dissociation reaction of $CH_3NH_3^+$ will be:

$CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)$

The equilibrium expression of $CH_3NH_3^+$ will be:

$K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}$

3 0

3 years ago

A chemist wants to make 6.5 L of a .350M CaCl2 solution. What mass of CaCl2(in g) should the chemist use?

kotykmax [81]

First M stands for Molarity which is (moles of solute) / (Liters of solution). we also know that moles = (mass) / (molar mass). so we can form some equations here. We know:
Molarity (M) = moles (mol) / Liters (L)
moles (mol) = (mass) / (molar mass)

we can substitute the (mass) / (molar mass) for (moles) and get:
M = [(mass) / (molar mass)] / Liters

we can now isolate mass and get
M * Liters * molar mass = mass

now we need to find the molar mass of CaCl2 which is 110.98 g/mol

plug the values in and get
.350M * 6.5L * 110.98 g/mol = mass

mass = 252.4795g however the 6.5L has only 2 sig figs so i would say

mass CaCl2 = 2.5 * 10 ^2 g

5 0

3 years ago

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Cobalt-63 has a half-life of 5.3 years. If a pellet that has been in storage for 15.9 years contains 40.0g of Cobalt-63, how muc

Cerrena [4.2K]

Answer:

320 g

Step-by-step explanation:

The half-life of Co-63 (5.3 yr) is the time it takes for half of it to decay.

After one half-life, half (50 %) of the original amount will remain.

After a second half-life, half of that amount (25 %) will remain, and so on.

We can construct a table as follows:

No. of Fraction Mass

half-lives t/yr Remaining Remaining/g

0 0 1

1 5.3 ½

2 10.6 ¼

3 15.9 ⅛ 40.0

4 21.2 ¹/₁₆

We see that 40.0 g remain after three half-lives.

This is one-eighth of the original mass.

The mass of the original sample was 8 × 40 g = 320 g

5 0

3 years ago