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Alex787 [66]
3 years ago
15

How did scientists find out about atoms before technology?

Chemistry
1 answer:
sasho [114]3 years ago
6 0

Answer:

There are three ways that scientists have proved that these sub-atomic particles exist. They are direct observation, indirect observation or inferred presence and predictions from theory or conjecture. Scientists in the 1800's were able to infer a lot about the sub-atomic world from chemistry.

Explanation:

Hope this helps

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In a reaction of a potential new fuel, it is found that when 2.81 moles of the fuel combusts, 1,612 kJ of energy is released. Wh
Tema [17]

Answer:

-573.67

Explanation:

whenever energy is released in a chemical reaction, we would then expect the delta H of the reaction to be negative because the reaction is an exothermic reaction.

now we have that 2.81 moles of fuel when it combusts would releases 1612kJ of energy

thus, 1 mole will release 1612/2.81 = -573.67kJ of heat

Therefore the delta H of the reaction = -573.67 kJ/mol

3 0
3 years ago
Consider the following chemical reaction:
laiz [17]

Answer:

B. 1.65 L

Explanation:

Step 1: Write the balanced equation

2 SO₂(g) + O₂(g) ⇒ 2 SO₃(g)

Step 2: Calculate the moles of SO₂

The pressure of the gas is 1.20 atm and the temperature 25 °C (298 K). We can calculate the moles using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.20 atm × 1.50 L / (0.0821 atm.L/mol.K) × 298 K = 0.0736 mol

Step 3: Calculate the moles of SO₃ produced

0.0736 mol SO₂ × 2 mol SO₃/2 mol SO₂ = 0.0736 mol SO₃

Step 4: Calculate the volume occupied by 0.0736 moles of SO₃ at STP

At STP, 1 mole of an ideal gas occupies 22.4 L.

0.0736 mol × 22.4 L/1 mol = 1.65 L

6 0
3 years ago
Calculate the change in entropy when 10.0 g of CO2 isothermally expands from a volume of 6.15 L to 11.5 L. Assume that the gas b
USPshnik [31]

Answer:

The change in entropy of the carbon dioxide is 1.183\times 10^{-3} kilojoules per Kelvin.

Explanation:

By assuming that carbon dioxide behaves ideally, the change in entropy (\Delta S), measured in kilojoules per Kelvin, is defined by the following expression:

\Delta S = m\cdot \bar c_{v}\cdot \ln \frac{T_{f}}{T_{o}}+m\cdot \frac{R_{u}}{M}\cdot \ln \frac{V_{f}}{V_{o}} (1)

Where:

m - Mass of the gas, measured in kilograms.

\bar c_{v} - Isochoric specific heat of the gas, measured in kilojoules per kilogram-Kelvin.

T_{o}, T_{f} - Initial and final temperatures of the gas, measured in Kelvin.

V_{o}, V_{f} - Initial and final volumes of the gas, measured in liters.

R_{u} - Ideal gas constant, measured in kilopascal-cubic meter per kilomole-Kelvin.

M - Molar mass, measured in kilograms per kilomole.

If we know that T_{o} = T_{f}, m = 0.010\,kg, R_{u} = 8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K}, M = 44.010\,\frac{kg}{kmol}, V_{o} = 6.15\,L and V_{f} = 11.5\,L, then the change in entropy of the carbon dioxide is:

\Delta S = \left[\frac{ (0.010\,kg)\cdot \left(8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)}{44.010\,\frac{kg}{kmol} } \right]\cdot \ln \left(\frac{11.5\,L}{6.15\,L}\right)

\Delta S = 1.183\times 10^{-3}\,\frac{kJ}{K}

The change in entropy of the carbon dioxide is 1.183\times 10^{-3} kilojoules per Kelvin.

4 0
3 years ago
Determine which compound would raise the boiling point of water the most
Olenka [21]
All of the provided salts will increase the boiling point of water due to Boiling point elevation is considered as one of the colligative properties that only affected by the number of particles present in solution regardless their nature.

All of the choices are considered as electrolytes (ionizes in water) so the one which gives higher number of ions will raise the boiling point the most so the correct answer is A) AlCl₃ due to it ionizes to give 4 ions one Al and 3 Cl while NaCl gives only 2 ions, CaCl₂ gives 3 ions and KBr gives 2 ions 

So the correct answer is A) 2.0 M AlCl₃ 
8 0
4 years ago
ammonia (NH3(g) Hf=-45.9 kJ/mol) reacts. with oxygen to produce nitrogen and water (H2O(g) Hf = -241.8 kJ/mol according to the e
kherson [118]

Answer:

ΔH°_rxn = -195.9 kJ·mol⁻¹

Explanation:

                              4NH₃(g) + 3O₂(g) ⟶ 2N₂(g) +6H₂O(g)

ΔH°_f/(kJ·mol⁻¹):    -45.9          0                 0        -241.8

The formula relating ΔH°_rxn and enthalpies of formation (ΔH°_f) is

ΔH°_rxn = ΣΔH°_f(products) – ΣΔH°_f(reactants)

ΣΔH°_f(products) = -6(241.8) = -1450.8 kJ

ΣΔH°_f(reactants) = -4(45.9) = -183.6 kJ

ΔH°_rxn =  (-1450.8 + 183.6) kJ = -1267.2 kJ

6 0
3 years ago
Read 2 more answers
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