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sergij07 [2.7K]

2 years ago

12

1. Which phase change is accompanied by the release of heat? A) H20(s)--> H2O(g) B) H20(l) ->H2O(s) C) H20(l)→ H2O(g) D) H

2O(s) -> H20(l)

1 answer:

docker41 [41]2 years ago

5 0

Answer:

B, liquid to solid.

Explanation: Since heat is being released, the particles for H2O would clump up. Heat is basically being taken out.

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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

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
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>

Set up: $\displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})$
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

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2 years ago

imine formation is an acid-catalyzed process, however the procedure performed in this does not require the addition of an extern

SVETLANKA909090 [29]

Answer:

See explaination

Explanation:

Reaction between p- toluidine and o- vaniline to form imine.This reaction does occur in the presence of a acidic source.

Here the -OH group in the o- vaniline is quite acidic so it acts a internal acidic catalyst. Hence the reaction occurred in the absence of any external acidic source.The reaction mechanism is given at the attached file.

please kindly see attachment.

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igomit [66]

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2 years ago

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yKpoI14uk [10]

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3 years ago

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