How does an increase in reactant concentration affect the rate of reaction?

Brianliest for whoever gets this right with step by step method

Sedaia [141]

The Mr is the mass numbers of each element added up so…. Fe = 56, O=16, H=1 … now add these up with the number of each element -> there’s 1 Fe, and 3 Os and 3 Hs as they are in brackets with a 3 outside-> (56+16+16+16+1+1+1=107) … your answer is 107

8 0

2 years ago

Draw the structure of triethylamine, which has the formula n(c2h5)3.

NARA [144]

I hope this is right

3 0

2 years ago

Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re

borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

A + B ⇄ C

where;

numbers of moles = 0.386 mol C (g)

Volume = 7.29 L

Molar concentration of C = $\frac{0.386}{7.29}$

= 0.053 M

A + B ⇄ C

Initial 0 0 0.530

Change +x +x - x

Equilibrium x x (0.0530 - x)

$K = \frac{[C]}{[A][B]}$

where

K is given as ; 78.2 atm-1.

So, we have:

$78.2=\frac{[0.0530-x]}{[x][x]}$

$78.2= \frac{(0.0530-x)}{(x^2)}$

$78.2x^2= 0.0530-x$

$78.2x^2+x-0.0530=0$

Using quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where; a = 78.2 ; b = 1 ; c= - 0.0530

= $\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

= $\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$ or $\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$

= 0.0204 or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

= 0.0530 - 0.0204

= 0.0326

Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

$V = \frac{nRT}{P}$

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K (fixed constant temperature )

V (volume) = ???

$V=\frac{(0.0734*0.0821*273.15)}{(1.00)}$

V = 1.64604

V ≅ 1.65 L

3 0

2 years ago

An iron bar at 200c is placed in thermal contact with an identical iron bar at 120c in an isolated system

Step2247 [10]

Conduction: In the conduction, the heat is transferred from the hotter body to the colder body until the temperature on both bodies are equal.

In thermal equilibrium, there is no heat transfer as the heat is transferred till the temperature on the bodies are not same.

In the given problem, an iron bar at 200°C is placed in thermal contact with an identical iron bar at 120°C in an isolated system. After 30 minutes, the thermal equilibrium is attained. Then, the temperature on both iron bars are equal.Both iron bars are at 160°C in an isolated system.

But in an open system, the temperatures of the iron bars after 30 minutes would be less than 160°C. There will be heat lost to the surrounding. The room temperature is 25°C. There will be exchange of the heat occur between the iron bars and the surrounding. But It would take more than 30 minutes for both iron bars to reach 160°C because heat would be transferred less efficiently.

7 0

2 years ago

Read 2 more answers

What is the mass in grams of 0.7350 moles of sodium?

Alik [6]

Answer:

Explanation:

1 mol of sodium = 23 grams (use the number on your periodic table).

0.7350 mol sodium = x

Cross multiply

1*x = 0.7350 * 23

x = 16.905

You will get slightly less than this, depending on your periodic table. But the method will be the same.

4 0

2 years ago