When we have stomach disorder, we take antacid to solve the problem. How does it help? Support your answer with an equation.

Let us assume that fe(oh)2(s) is completely insoluble, which signifies that the precipitation reaction with naoh(aq) (presented

Yuki888 [10]

17.8 mL NaOH

<em>Step 1.</em> Write the chemical equation

Fe^(2+) + 2NaOH → Fe(OH)2 + 2Na^(+)

<em>Step 2.</em> Calculate the moles of Fe^(2+)

Moles of Fe^(2+) = 500 mL Fe^(2+) × [0.0230 mmol Fe^(2+)]/[1 mL Fe^(2+)]

= 11.50 mmol Fe^(2+)

<em>Step 3.</em> Calculate the moles of NaOH

Moles of NaOH = 11.50 mmol Fe^(2+) × [2 mmol NaOH]/[1 mmol Fe^(2+)]

= 23.00 mmol NaOH

<em>Step 4.</em> Calculate the volume of NaOH

Volume of NaOH = 23.00 mmol NaOH × (1 mL NaOH/1.29 mmol NaOH)

= 17.8 mL NaOH

3 0

3 years ago

Rahul and Sonia were playing with blocks and each of them made a train out of them. Both of them thought of measuring the length

dalvyx [7]

Answer:

See explanation

Explanation:

(a) The handspan is the distance between the tip of the thumb and tip of the little finger when the hand is fully stretched.

b) It is impossible for them to obtain an accurate result by the use of a handspan because it doesn't use any standard unit for measurement. It is entirely subjective and highly prone to human errors.

c) Anupam is quite intelligent and active

3 0

3 years ago

SpongeBob & Friends and the Scientific Method

yarga [219]

Answer:

1. Which people are in the control group? The people who received the mint without the secret ingredient

(Group B) would be the control group.

2. What is the independent variable? Secret ingredient in the breath mint

3. What is the dependent variable? Amount of breath odor (or bad breath)

4. What should Mr. Krabs’ conclusion be? The breath mint with the secret ingredient appears to reduce the

amount of breath odor more than half the time, but it is not 100% effective.

5. Why do you think 10 people in group B reported fresher breath? This may be due to the placebo effect.

6 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

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: $\displaystyle 0.551373 \ g \ NaCl$