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Margaret [11]
3 years ago
5

What mass of hydrochloric acid (in grams) can 2.7 g of sodium bicarbonate neutralize? (Hint: Begin by writing a balanced equatio

n for the reaction between aqueous sodium bicarbonate and aqueous hydrochloric acid.)
Chemistry
1 answer:
Julli [10]3 years ago
3 0

Answer:

1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate

Explanation:

Step 1: Data given

Mass of sodium bicarbonate = 2.7 grams

Step 2: The balanced equation

HCl + NaHCO3 ⇔  NaCl + H2O + CO2

Step 3: Calculate moles NaHCO3

moles NaHCO3 =2.7 g / 84 g/mol= 0.032 moles

Step 4: Calculate moles HCl

For 1 mol NaHCO3 we need 1 mol HCl

For 0.032 moles NaHCO3 = 0.032 moles HCl

Step 5: Calculate mass HCl

Mass HCl = moles HCl * molar mass HCl

mass HCl = 0.032 * 36.46 g/mol= 1.17 grams

1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate

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Answer:

it binds molecules like a chemical bond-breaking

Explanation:

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3 0
3 years ago
How isa molecule of mercury different from a molocue of water
Genrish500 [490]
<h2>Answer:</h2>

The density of mercury molecule is higher than water.

<h3>Explanation:</h3>

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8 0
3 years ago
A jogger runs a mile in 8.92 minutes. 1 mi=1609m; calculate her speed in km/hr
olchik [2.2K]

Answer:

Speed = 3.30 \frac{km}{hr}

Explanation:

Given

Distance = 1 mile

Time = 8.92 minutes

Required

Calculate Speed in km/hr

Speed is calculated as thus;

Speed = \frac{Distance}{Time}

Substitute 1 mile for distance and 8.92 minutes for time

Speed = \frac{1\ mile}{8.92\ minutes}

Convert Miles to Kilometres

If

1\ mile = 1609\ m;

Then

1\ mile = \frac{1609\ km}{1000}

1\ mile = 1.609\ km --- (1)

Convert minutes to hour

1\ minutes = 0.0167\ hour

Multiply both sides by 8.92

8.92 * 1\ minutes = 0.0167\ hour * 8.92

8.92 \ minutes = 0.148964\ hour ---- (2)

By substituting (1) and (2) in Speed = \frac{1\ mile}{8.92\ minutes}, we have

Speed = \frac{1.609\ km}{0.48694\ hour}

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The molecular model tells us how the atoms/elements are bonded together.

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Molecular model:   H -- C -- H
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<span>                                  H</span>
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