Which activity demonstrates chemical weathering?

What is the identity of an ion having 46 protons, 42 electrons and 60 neutrons?

Anon25 [30]

Answer:

first u lick it then out some sliva on it then rub it and once the volcano erupts u put it back in ur mouth and blow real hard ok!

Explanation:

repeat this process till the dong bleeds!

3 0

2 years ago

Chem. Assignment <br><br> I need help...with answers 1-6 thanks

Anna11 [10]

Answer:

1. an educated guess

2. data

3. what changes in experiment

4. what stays the same in both groups

5. the group where nothing changes, normal

6. group with independent variable, what's being tested

8 0

3 years ago

Be sure to answer all parts. Express the rate of reaction in terms of the change in concentration of each of the reactants and p

Tanzania [10]

Answer : The [H] is increasing at the rate of 0.36 mol/L.s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}$

$\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}$

$\text{Rate of disappearance of }F=-\frac{d[F]}{dt}$

$\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}$

$\text{Rate of formation of }H=+\frac{d[H]}{dt}$

$\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}$

Given:

$-\frac{d[D]}{dt}=0.18mol/L.s$

As,

$-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s$

and,

$+\frac{d[H]}{dt}=2\times 0.18mol/L.s$

$+\frac{d[H]}{dt}=0.36mol/L.s$

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

5 0

3 years ago

Which of the following is true regarding nuclear fission and nuclear fusion?

Zarrin [17]

<h2>Nuclear Fission and Nuclear Fusion - Option C</h2>

Nuclear fission and nuclear fusion both of these processes can provide energy. Nuclear fission is the process in which heavy nucleus splits into smaller parts. When they split into smaller particles then it releases energy.

On the other hand, nuclear fusion is the process in which small particles fuse together to form a heavy nucleus. With the formation of heavy nucleus, it also provides energy.

Therefore, both these processes release or provide energy.

3 0

3 years ago

Read 2 more answers

Part 1: What is the final volume in milliliters when 0.730 L of a 44.8 % (m/v) solution is diluted to 23.3 % (m/v)?

Andre45 [30]

part 1 : the final volume : 1.404 L

part 2 : the initial concentration : 4.06 M

<h3>Further explanation </h3>

Dilution is the process of adding a solvent to get a more dilute solution.

The moles(n) before and after dilution are the same.

Can be formulated :

M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution

V₁ = volume of the solution before dilution

M₂ = Molarity of the solution after dilution

V₂ = Molarity volume of the solution after dilution

part 1 :

M₁=44.8%

V₁=0.73 L

M₂=23.3%

$\tt V_2=\dfrac{M_1.V_1}{M_2}\\\\V_2=\dfrac{44.8\times 0.73}{23.3}\\\\V_2=1.404~L$

part 2 :

V₁=739 ml=0.739 L

V₂=1.5 L

M₂=2

$\tt M_1=\dfrac{M_2.V_2}{V_1}\\\\M_1=\dfrac{2\times 1.5}{0.739}\\\\M_1=4.06$

6 0

3 years ago