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JulsSmile [24]
4 years ago
5

The equation below shows the products formed when a solution of silver nitrate (AgNO3) reacts with a solution of sodium chloride

(NaCl).
AgNO3 + NaCl → NaNO3 + AgCl

(Reactants) → (Products)


In this equation, the total mass of the reactants is
A) equal to the mass of AgCl.
B) less than the mass of AgCl.
C) equal to the total mass of the products.
D) greater than the total mass of the products.
Chemistry
2 answers:
Alecsey [184]4 years ago
4 0
C<span>) <span>igual a la masa total de los productos. <span> equal to the total mass of the products.

</span></span></span>In this equation, the total mass of the reactants is <span>equal to the total mass of the products.</span> This is in accordance with the law of conservation of mass.
d1i1m1o1n [39]4 years ago
3 0
I would say that the answer has to be C
Since there is no change in mols on both sides of the equation the mass is constant
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What are heterogeneous and homogeneous?
vichka [17]
A homogeneous mixture has the same uniform appearance and composition throughout.
A heterogeneous mixture consists of visibly different substances or phases. The three phases or states of matter are gas, liquid, and solid.
A homogeneous mixture has the same uniform appearance and composition throughout. Many homogeneous mixtures are commonly referred to as solutions. A heterogeneous mixture consists of visibly different substances or phases.
3 0
3 years ago
Given 24.3 g of LICI, Find moles of PbCI, produced.<br> PbSO4 + LiCl → PbCl2 + Li2SO4
Tema [17]

Answer:

0.29 moles of PbCl₂

Explanation:

Given data:

Mass of lithium chloride = 24.3 g

Moles of PbCl₂ = ?

Solution:

Chemical equation;

PbSO₄ + 2LiCl    →    PbCl₂ + Li₂SO₄

Number of moles of LiCl:

Number of moles = mass/ molar mass

Number of moles = 24.3 g/ 42.394 g/mol

Number of moles = 0.57 mol

Now we will compare the moles of PbCl₂ with LiCl .

            LiCl               :          PbCl₂

              2                 :            1

              0.57            :        1/2×0.57 = 0.29 mol

4 0
3 years ago
What is the value of for this aqueous reaction at 298 k? a b↽−−⇀c dδ°=20. 46 kj/mol
sineoko [7]

The value for this aqueous reaction at 298 k? a b↽−−⇀c dδ°=20. 46 KJ/mol is 9.91 mol. in equilibrium

<h3>What is an aqueous reaction in equilibrium?</h3>

When a chemical reaction happens at the liquid state and the formation of reactant and product is the same then the reaction is known as an aqueous reaction in equilibrium denoted by K.

δG = − R T ln

          R = universal gas constant 8.313

          δG= 20. 46 kj/mol

           T =  298 k or 24.4 in celcius.

substituting the value in the equation.

20. 46 kj/mol = 8.313 × 24.4 in celcius × K

K =  8.313 × 24.4 in celcius / 20. 46 kj/mo

k = 9.91 mol .

Therefore, The value of this aqueous reaction at 298 k? a b↽−−⇀c dδ°=20. 46 KJ/mol is 9.91 mol. in equilibrium

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5 0
2 years ago
Calculate each of the following:<br> (b) Mass % of Mn in potassium permanganate
Brilliant_brown [7]

Mass percent is the estimation of the mass of the element in a compound in percentage form. The mass % of manganese (Mn) in potassium permanganate is 34.76 %.

<h3>What is the mass percentage?</h3>

The mass percentage is defined by dividing the mass of the element by the molecular mass of the compound followed by multiplying it by 100%. It is given as,

Mass % =  (Atomic Mass of element ÷ Molecular Mass) × 100

As it is known that,

Atomic Mass of manganese =  54.94 g/mol

Molecular Mass of potassium permanganate =  158.034 g/mol

Substituting values to calculate mass % as:

% Mn  =  (Atomic Mass of Mn ÷ Molecular Mass of KMnO₄) × 100

= (54.94 g/mol ÷ 158.034 g/mol) × 100

= 34.76 %

Therefore, 34.76% Mn is present in KMnO₄.

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3 0
2 years ago
Compare and contrast complete equations and complete ionic equations
olga55 [171]

Answer:

Explanation: The key difference between Complete ionic and net ionic equation is that complete ionic equation gives all the ionic species participated in the chemical reaction whereas net ionic reaction gives the chemical species participated in the formation of the final product.

7 0
3 years ago
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