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4 years ago

What is the empirical formula for Hg2(NO3)2

Chemistry

1 answer:

Fynjy0 [20]4 years ago

6 0

HgNO3 The subscripts on Hg and NO3 are not divisible by 2

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geniusboy [140]

Answer:

B. because there is two equations just like commutative property in math its the same thingish

Explanation:

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3 years ago

Someone help me out with this brain pop ⚠️

erik [133]

The answer is b because if you’re fan is blowing and you spend it the opposite way, the air will be cooler

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3 years ago

PLEASE HELP IM TAKING A TEST!!! Consider the reaction 4FeS2 + 11O2 → 2Fe2O3 + 8SO2. If 8 moles of FeS2 react with 15 moles of O2

chubhunter [2.5K]

Answer:

c-)O2

Explanation:

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2 years ago

I neeeed helppppp!!!!!!!

Nezavi [6.7K]

Answer:

The first claim is false. Natural resources can most definitely exist without synthetic products.

Explanation:

All synthetic products are derived from natural resources as everything has to come from a source in the beginning. However, most natural resources themselves are the source, make sense? For example, water is a naturally occurring resource, but even water is derived from specific components, however it doesn't need anything synthetic to exist. Now take hairspray, to make hairspray you need water and synthetic resources, see where I'm going with that?

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3 years ago

You use an analytical balance to measure the mass of a weigh boat and record a mass of 1.5624 g. You then add some NaHCO3 to the

never [62]

Answer:

The minimum number of grams of acetic acid that we would need to fully react with your NaHCO3, is 0.263 grams.

Explanation:

Step 1: Data given

Mass weigh boat = 1.5624 grams

Mass of boat + NaHCO3 = 1.92...

Step 2: Calculate ranges

The minimum possible mass of boat + NaHCO3 = 1.9200g

Minimum mass of NaHCO3 = 1.9200g - 1.5624g = 0.3576g

The maximum possible mass of boat + NaHCO3 = 1.9299g

Maximum mass of NaHCO3 = 1.9299g - 1.5624g = 0.3675g

What is the minimum mass of acetic acid that must be used to react with all the NaHCO3 possible in the boat:

This means what mass of CH3COOH reacts with 0.3675g NaHCO3

NaHCO3 + CH3COOH → CH3COONa + CO2 + H2O

1 mole of NaHCO3 consumed, needs 1 mole CH3COOH to produce 1 mole of CH3COONa, 1 mole CO2 and 1 mole H2O

Step 3: Calculate moles NaHCO3

Moles NaHCO3 = mass NaHCO3 / Molar mass NaHCO3

moles NaHCO3 = 0.3675 grams / 84g/mol

moles NaHCO3 = 0.004375 moles

Since 1 mole of NaHCO3 consumed, needs 1 mole CH3COOH

For 0.004375 moles NaHCO3, we need 0.004375 moles CH3COOH

Step 4: Calculate mass of CH3COOH

mass of CH3COOH = moles CH3COOH * Molar mass CH3COOH

mass of CH3COOH = 0.004375 moles * 60.05 g/mol

mass CH3COOH = 0.2627 grams ≈ 0.263 grams

The minimum number of grams of acetic acid that we would need to fully react with your NaHCO3, is 0.263 grams.

6 0

3 years ago