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mamaluj [8]
2 years ago
12

Would be grateful if someone could show the process for solving this problem.

Chemistry
1 answer:
In-s [12.5K]2 years ago
4 0
I don’t know i’m trying to get some points
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The following reaction shows calcium chloride reacting with silver nitrate.
adell [148]

Grams of Ca(NO₃)₂ produced : 0.985 g

<h3>Further explanation</h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

CaCl₂ + 2AgNO₃ → 2AgCl + Ca(NO₃)₂

MW AgNO₃ : 107.9+14+3.16=169.9

mol AgNO₃ :

\tt \dfrac{2}{169.9}=0.012

mol ratio Ca(NO₃)₂ : AgNO₃ = 1 : 2, so mol Ca(NO₃)₂ :

\tt \dfrac{1}{2}\times 0.012=0.006

MW Ca(NO₃)₂ : 40.1+2.14+6.16=164.1 g/mol

mass Ca(NO₃)₂ :

\tt 0.006\times 164.1=0.985

7 0
3 years ago
At which pesan pressure lowest The image shows a representation of mountains of various heights, numbered 1, 2, and 3 The plami
ira [324]

1 is the lowest pressure.........

4 0
2 years ago
5. Write the names of the following ions.
Talja [164]
A.) Phosphate ion or Orthophosphate
d.) Hydroxide
D.) Ammonium
e.) Iron
C.) Nitrate
f.) Sulfur dioxide
5 0
2 years ago
A sample of iron with a mass of 200g releases 9840 cal when it freezes at it’s freezing point. What is the molar heat of fusion
antoniya [11.8K]

Answer:-

2747.7 Cal mol -1

Explanation:-

Molar heat of Fusion is defined as the amount of heat necessary to melt (or freeze) 1 mole of a substance at its melting point.

Atomic mass of Iron = 55.845 g mol-1

Mass of Iron = 200 g

Number of moles of Iron = 200 g / (55.845 g mol-)

= 3.581 moles

Heat released = 9840 Cal

Molar heat of Fusion = Heat released / Number of moles

= 9840 Cal / 3.581 moles

= 2747.7 Cal mol -1

8 0
2 years ago
Read 2 more answers
How many moles are in 3.113 g of Au?Molar mass of Au=197 g/mol
SVEN [57.7K]
<h3>Answer:</h3>

0.0157 g Au

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

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis

<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
  • Left to Right
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.113 g Au

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

Molar Mass of Au - 197.87 g/mol

<u>Step 3: Convert</u>

<u />3.113 \ g \ Au(\frac{1 \ mol \ Au}{197.87 \ g \ Au} ) = 0.015733 g Au

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

0.015733 g Au ≈ 0.0157 g Au

3 0
2 years ago
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