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3 years ago

The mass and charge of all three sub atomic particles

Chemistry

2 answers:

Paladinen [302]3 years ago

5 0

Answer:

Protons, neutrons, and electrons are the three main subatomic particles found in an atom. Protons have a positive (+) charge. An easy way to remember this is to remember that both proton and positive start with the letter "P." Neutrons have no electrical charge.

Andre45 [30]3 years ago

5 0

Answer:

proton mass 1 charge +1

neutron mass 1 no charge

electron no mass charge -1

Explanation:

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liberstina [14]

Answer:

The dissociation constant for hydrochloric acid is greater than the dissociation constant for acetic acid

Explanation:

answer is correct

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3 years ago

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emmasim [6.3K]

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3 years ago

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It takes 42.25 ml of 0.0500 M Na2S2O3 solution to completely react with the iodine present in a 150.0 ml iodine solution. How ma

svlad2 [7]

Answer:

There were 0.268 grams I2 present

Explanation:

Step 1: Data given

Volume of Na2SO3 = 42.25 mL = 0.04225 L

Molarity of Na2SO3 = 0.0500 M

Volume of iodine = 150.0 mL

Step 2: The balanced equation

I2 + 2 Na2SO3 → Na2S2O6 + 2 NaI

Step 3: Calculate moles of Na2SO3

Moles Na2SO3 = molarity * volume

Moles Na2SO3 = 0.0500 M * 0.04225 L

Moles Na2SO3 = 0.0021125 moles

Step 4: Calculate moles I2

For 1 mol I2 we need 2 moles Na2SO3

For 0.0021125 moles Na2SO3 we have 0.00105625

Step 5: Calculate mass of I2

Mass I2 = moles * molar mass

Mass I2 = 0.00105625 * 253.8 g/mol

Mass I2 = 0.268 grams

There were 0.268 grams I2 present

3 0

4 years ago

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A weather balloon is inflated to a volume of 2.2 x 10^3 L with 37.4 g of helium. What is the density of helium in grams per lite

Ede4ka [16]

Density = mass/volume
substituting the given values :
We get :
D = 37.4 / 2.2 x 10^3
D = 0.017 grams/liter

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3 years ago

Calculate ∆h0 for the reaction 2 n2(g) + 5 o2(g) −→ 2 n2o5(g) given the data h2(g) + 1 2 o2(g) −→ h2o(ℓ) ∆h 0 f = −283.7 kj/mol

8_murik_8 [283]

Based on Hess's Law:

2 N2(g) + 6 O2(g) + 2 H2(g) −→ 4 HNO3(l) ∆Hf = (−171.9 kJ/mol)(4 mol) 
2 H2O(l) −→ 2 H2(g) + O2(g) ∆Hf = (-283.8 kJ/mol)(2 mol)(-1) →times -1, rxn is reversed 
4 HNO3(l)−→ 2 N2O5(g) + 2 H2O(l) ∆Hf = (-76.4 kJ/mol)(2 mol)(-1) →times -1, rxn is reversed 

2 N2(g) + 5 O2(g) −→ 2 N2O5(g) ∆H0 = 32.8 kJ

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4 years ago