Calculate the empirical formula of a compound with the following

netineya [11]

Here is 5

Dissolved Load - elements dissolved in solution

Suspended Load - very fine grained sediment such as clay and silt carried in suspension. The size grains that can be carried in suspension are dependent on the current velocity

Wash Load - a subset of the suspension load, extremely small particles (clay) that will remain in suspension independent of turbulence in the river

Saltation Load - particles that are temporarily carried in suspension but move by bouncing along the bottom

Bed Load - sediment that moves by rolling or sliding along the bottom. These are generally the coarser grained sediments such as sand and gravel.

5 0

3 years ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

3 years ago

A truck traveled at a constant speed for 6 hours. Then, it

gavmur [86]

Answer:

2.4 hrs

Explanation:

The constant speed of the truck for 6 hrs can be calculated by: speed=distance/time. Speed =(876-228)/6=648/6=108m/s. So the decreased speed = (108-13)=95m/h. Now, speed =distance /time We get 95m/h = 228/t. t=228/95 hrs = 2.4 hrs PLEASE MARK ME THE BRAINLIEST!!

7 0

3 years ago

Read 2 more answers

How many moles of Na2SO4, are produced from 7.00 moles of (NH4)2SO4?

Harlamova29_29 [7]

Answer:

dumb stupid restarted illiterate STUPID students ARE JUST GOING TO DIE

8 0

3 years ago

Classify each of the following reactants and products as an acid or base according to the Bronsted theory: H3PO4 + C6H5O- <=&

Alexxandr [17]

Answer:

H₃PO₄ is an acid because donates the proton to fenolate.

Fenolate is the base because accepts the proton from the acid.

Explanation:

Bronsted theory mentioned that acid is the one that donates a proton to another compound and base is the one that receives it.

H₃PO₄ + C₆H₅O⁻ ⇄ H₂PO₄⁻ + C₆H₅OH

acid base conj. base conj. acid

H₃PO₄ is an acid because donates the proton to fenolate.

Fenolate is the base because accepts the proton from the acid.

If we follow the dissociation, the diacid phosphate can donate two more protons, it is still a Bronsted acid, but it can act as an acid or a base. This is called amphoteric.

6 0

3 years ago