What is the molecular weight of one mole of H₂CO3?

How does marketing affect the consumer mindset?

RoseWind [281]

Answer:

Marketing affects the consumer mindset by leaving them with the decision of either purchasing a product or not and why it should be purchased or not.

6 0

3 years ago

What information can be learned from bloodstains at a crime scene?

tangare [24]

you can learn who's blood it is, you can also see the blood patterns and how the crime maybe happened.

3 0

4 years ago

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In

qaws [65]

The question is incomplete, here is the complete question:

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In the first step, calcium carbide and water react to form acetylene and calcium hydroxide:

$CaC_2(s)+2H_2O(g)\rightarrow C_2H_2(g)+Ca(OH)_2(s)$

In the second step, acetylene, carbon dioxide and water react to form acrylic acid:

$6C_2H_2(g)+3CO_2(g)+4H_2O(g)\rightarrow 5CH_2CHCO_2H(g)$

Write the net chemical equation for the production of acrylic acid from calcium carbide, water and carbon dioxide. Be sure your equation is balanced.

Answer: The net chemical equation is written below.

Explanation:

The intermediate balanced chemical reaction are:

(1) $CaC_2(s)+2H_2O(g)\rightarrow C_2H_2(g)+Ca(OH)_2(s)$ ( × 6 )

(2) $6C_2H_2(g)+3CO_2(g)+4H_2O(g)\rightarrow 5CH_2CHCO_2H(g)$

To omit acetylene from the net chemical reaction, we multiply Equation (1) by 6.

Equation 1: $6CaC_2(s)+12H_2O(g)\rightarrow 6C_2H_2(g)+6Ca(OH)_2(s)$

Equation 2: $6C_2H_2(g)+3CO_2(g)+4H_2O(g)\rightarrow 5CH_2CHCO_2H(g)$

Net chemical equation: $6CaC_2(s)+3CO_2(g)+16H_2O(g)\rightarrow 5CH_2CHCO_2H(g)+6Ca(OH)_2(s)$

Hence, the net chemical equation is written above.

6 0

3 years ago

At a certain temperature, this reaction establishes an equilibrium with the given equilibrium constant, Kc. 3 A ( g ) + 2 B ( g

cricket20 [7]

Answer:

The concentrations of A, B, and C at equilibrium:

[A] = 0.0 M

[B] = 2.7 M

[C] = 2.4 M

Explanation:

Concentration of 1.80 mol of A in 1.00 L container :

$[A]=\frac{1.80 mol}{1.00 L}=1.80 M$

Concentration of 3.90 mol of B in 1.00 L container :

$[B]=\frac{3.90 mol}{1.00 L}=3.90 M$

$3A(g)+2B(g)\rightleftharpoons 4C(g), K_c=2.13\times 10^{17}$

Initially

1.80 M 3.90 M 0

At equilibrium

(1.80-3x)M (3.90-2x) 4x

The expression of an equilibrium constant is given by :

$K_c=\frac{[C]^4}{[A]^3[B]^2}$

$2.13\times 10^{17}=\frac{(4x)^4}{(1.80-3x)^3\times (3.90-2x)^2}$

Solving for x:

x = 0.600

The concentrations of A, B, and C at equilibrium:

[A] = [1.80-3x]=[1.80-3 × 0.600]= 0 M

[B] = [3.90-2x] = [3.90-2 × 0.600] = 2.7 M

[C] = [4x] =[4 × 0.600 M] = 2.4 M

6 0

3 years ago

Rank the elements b, be, n, c, and o in order of increasing first ionization energy. (use the appropriate <, =, or > symbo

Dafna11 [192]

Be, B, C, N (going from smallest to biggest)

6 0

3 years ago