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tester [92]
3 years ago
5

If I added 50 mL of a 1.0 M HCIO2 solution to 50 mL of a 1.0 M NaOH

Chemistry
1 answer:
kozerog [31]3 years ago
3 0

Answer:

pH=7

Explanation:

0.05 L*1.0M= 0.05 mol HClO2  and 0.05 mol NaOH      

HClO2 +     NaOH ---> NaClO2 + H2O

0.05 mol     0.05 mol

As we can see acid and base  react completely,  solution will be neutral, so pH =7

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If copper carbonate system is heated for too short a time, what will you notice for mass percent of copper be too large or too s
sammy [17]

Answer:

The mass percent of copper as element is the same.

Explanation:

First of all we need the reaction that is presented below:

CuCO_{3 (s)} +Heat → CuO_{(s)} + CO_{2 (g)}

The mass percent of copper (Cu) as element is the same because of during the reaction the element only transform its nature from copper carbonate (CuCO_{3 (s)}) to copper oxide (CuO_{(s)}), the latter is a solid and will remain in the system.

On the other hand, you will note that the global percentage mass will be small because of the reaction produce (CO_{2 (g)}) that is a gas and this one will escape for the system.  

Have a great day!

3 0
3 years ago
Choose the correct coefficients for the equation? ____NaCI>____Na+___C12​
Wewaii [24]

Explanation:

Ca + NaCl ----> CaCl2 + Na

6 0
3 years ago
PLEASE ANSWER CORRECTLY FOLLOWING GUIDLINES SO ANSWER WONT GET DELETED I REALLY NEED HELP
blagie [28]

Chemical reaction

equation

reactants

products

yields

mass

balanced

atoms

coefficients

numbers

element

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substance

two

Explanation:

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8 0
2 years ago
The most common element in the atmosphere are
Ivanshal [37]
Nitrogen and oxygen are the most prevalent in the atmosphere.
8 0
3 years ago
1 Na2CO3(aq) + 1 CaCl2(aq) → 1 CaCO3(s) + 2 NaCl(aq) 4. Use the balanced chemical equation from the last question to solve this
LenKa [72]
<h3>Answer:</h3>

0.6 g NaCl

<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
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂

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

[RxN] Na₂CO₃ → 2NaCl

Molar Mass of Na - 22.99 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                    \displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})
  2. Multiply/Divide:                                                                                               \displaystyle 0.551373 \ g \ NaCl

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

0.551373 g NaCl ≈ 0.6 g NaCl

5 0
2 years ago
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