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Sidana [21]
3 years ago
12

if 115 G of a substance reacts with 84 grams of another substance what will be the mass of the products after the reaction​

Chemistry
1 answer:
jolli1 [7]3 years ago
8 0

Answer:

199 g

Explanation:

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

Explanation:

This law was given by french chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

For example:

In given reaction there are two reactant one with a mass of 115 g and other with the mass of 84 g thus the resultant product must have a mass of 199 g.

Chemical equation:

A + B → AB

115 g + 84 g = 199 g

119 g = 199 g

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3 years ago
Determine the molar solubility of pbso4 in pure water. ksp (pbso4) = 1.82 x 10-8.
Misha Larkins [42]
Use the ICE table approach as solution:

           PbSO₄   --> Pb²⁺ + SO₄²⁻
I             -                 0          0
C           -                +s         +s
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5 0
3 years ago
2c+02=2CO2. The moles of co2 produced when 0.25 moles of O2 react is?​
Sophie [7]
<h3>Answer:</h3>

\displaystyle 0.5 \ mol \ CO_2

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2C + O₂ → 2CO₂

[Given] 0.25 moles O₂

[Solve] moles CO₂

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol O₂ → 2 mol CO₂

<u>Step 3: Stoichiometry</u>

  1. [DA] Set up:                                                                                                     \displaystyle 0.25 \ moles \ O_2(\frac{2 \ mol \ CO_2}{1 \ mol \ O_2})
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2 years ago
Is oil and water a solution, colloid, or suspension?
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There are some data that suggest that zinc lozenges can significantly shorten the duration of a cold. If the solubility of zinc
zhuklara [117]

Answer:

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Explanation:

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Solubility product of Zn(CH_{3}COO)_{2} (K_{sp}) is written as-            K_{sp}=[Zn^{2+}][CH_{3}COO^{-}]^{2}

Where [Zn^{2+}] and [CH_{3}COO^{-}] represents equilibrium concentration (in molarity) of Zn^{2+} and CH_{3}COO^{-} respectively.

Molar mass of Zn(CH_{3}COO)_{2} = 183.48 g/mol

So, solubility of Zn(CH_{3}COO)_{2} = \frac{43.0}{183.48}M = 0.234M

1 mol of Zn(CH_{3}COO)_{2} gives 1 mol of Zn^{2+} and 2 moles of CH_{3}COO^{-} upon dissociation.

so,   [Zn^{2+}] = 0.234 M and [CH_{3}COO^{-}] = (2\times 0.234)M=0.468M

so, K_{sp}=(0.234)\times (0.468)^{2}=0.0513          

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