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3 years ago

What is commonly used to run a central heating system?

1 answer:

7 0

Answer :- Electrical energy

Explanation :-
Geothermal energy is too large scale
solar energy is inefficient
Nuclear does more harm

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How many grams of H2SO4 are needed to prepare 500. mL of a .250M solution?

zavuch27 [327]

Answer:

We need 12.26 grams H2SO4

Explanation:

Step 1: Data given

Volume of a H2SO4 solution = 500 mL = 0.500 L

Concentration of the H2SO4 solution = 0.250 M

Molar mass of H2SO4 = 98.08 g/mol

Step 2: Calculate moles H2SO4

Moles H2SO4 = concentration * volume

Moles H2SO4 = 0.250 M * 0.500 L

Moles H2SO4 = 0.125 moles

Step 3: Calculate mass of H2SO4

Mass of H2SO4 = moles * molar mass

Mass of H2SO4 = 0.125 moles * 98.08 g/mol

Mass of H2SO4 = 12.26 grams

We need 12.26 grams H2SO4

7 0

3 years ago

What do all nickel atoms have in common

Mama L [17]

All nickel atoms would have the same number of protons or atomic number.

4 0

3 years ago

A compound contains 72% magnesium and 28% nitrogen. What is its empirical formula?

SashulF [63]

So the empirical formula is Mg3N2

7 0

3 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

A sample of a compound is decomposed in the laboratory and produces 330 g carbon, 69.5 g hydrogen, and 220.2 g oxygen. Calculate

Zarrin [17]

Answer: Empirical formula is $C_2H_5O$

Explanation: We are given the masses of elements present in a sample of compound. To evaluate empirical formula, we will be following some steps.

Step 1 : Converting each of the given masses into their moles by dividing them by Molar masses.

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of Carbon = 12.0 g/mol

Molar mass of Hydrogen = 1.0 g/mol

Molar mass of Oxygen = 16.0 g/mol

Moles of Carbon = $\frac{330g}{12g/mol}=27.5moles$

Moles of Hydrogen = $\frac{69.5g}{1g/mol}=69.5moles$

Moles of Oxygen = $\frac{220.2g}{16g/mol}=13.76moles$

Step 2: Dividing each mole value by the smallest number of moles calculated above and rounding it off to the nearest whole number value

Smallest number of moles = 13.76 moles

$\text{Mole ratio of Carbon}=\frac{27.5moles}{13.76moles}=1.99\approx 2$

$\text{Mole ratio of Hydrogen}=\frac{69.5moles}{13.76moles}=5.05\approx 5$

$\text{Mole ratio of Oxygen}=\frac{13.76moles}{13.76moles}=1$

Step 3: Now, the moles ratio of the elements are represented by the subscripts in the empirical formula

Empirical formula becomes = $C_2H_5O$

7 0

3 years ago