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77julia77 [94]
3 years ago
6

270kJ are needed to keep a 75 watt light bulb burning for 1 hour. How many kilocalories are needed to burn this light bulb for 3

hours?
Chemistry
1 answer:
elena-s [515]3 years ago
8 0

Answer:

For 1 hour 75w light bulb requires 270 kj  for burning

for 3 hours 75 w light bulb requires 270*3 = 810kj for burning

Explanation:

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How many moles are present in 360 grams of water.
alexandr402 [8]

Answer:

From point, 1 mole of water = molar mass of water =18g 20 moles of water = 18 g x 20 = 360g (iv) From point, 6.022 x 1023 molecules of water = 1 mole = 18g of water 1.2044 x 1025 molecules of water Therefore, points (ii) and (iv) represent 360 g of water.

4 0
2 years ago
n-Butane (C4H10) is burned with stoichiometric amount of oxygen. Determine the mole fraction of carbon dioxide and water in the
Fudgin [204]

Answer:

See details below

Explanation:

The balanced reaction equation is given below:

2C_{4} H_{10} + 13O_{2} → 8CO_{2} + 10H_{2} O

Mole fraction of CO2 to H20

= 8/10 = \frac{4}{5}

Mole ratio of C4H10 to CO2 is 2:8 = 1:4

1 mole of n-butane - 38.12 g

4 moles - ?

= 152.48g fuel consumed.

8 0
3 years ago
Why is melting point considered a physical property
miskamm [114]

Answer:

Because you can physically see the object melting when it comes to the melting point. The objects texture, color, temperature, shape, and state of matter (solid, liquid, gas) are possibly changing.

6 0
3 years ago
The density of an element is 19.3 g/cm^3. What is its density in kg/m^3?
Karolina [17]
First, you need to know 1 kg = 10^3 g. And 1 m^3 = 10^6 m^3. So the 1 g/cm3 = 10^3 kg/m3. So the answer is 1.93*10^4 kg/m3.
5 0
3 years ago
A chemist fills a reaction vessel with mercurous chloride solid, mercury (I) aqueous solution, and chloride aqueous solution at
makvit [3.9K]

Answer:

ΔG° = -533.64 kJ

Explanation:

Let's consider the following reaction.

Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)

The standard Gibbs free energy (ΔG°) can be calculated using the following expression:

ΔG° = ∑np × ΔG°f(products) - ∑nr × ΔG°f(reactants)

where,

ni are the moles of reactants and products

ΔG°f(i) are the standard Gibbs free energies of formation of reactants and products

ΔG° = 1 mol × ΔG°f(Hg₂²⁺) + 2 mol × ΔG°f(Cl⁻) - 1 mol × ΔG°f(Hg₂Cl₂)

ΔG° = 1 mol × 148.85 kJ/mol + 2 mol × (-182.43 kJ/mol) - 1 mol × (-317.63 kJ/mol)

ΔG° = -533.64 kJ

3 0
3 years ago
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