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Lelechka [254]
2 years ago
13

Based on the chart, determine the identity of a substance that increases in temperature by 11.4oC when 1250J of energy are added

to a 55g sample.
Substance Specific Heat (J/g oC)
Lithium 3.56
Vegetable Oil 1.99
Air 1.02
Iron 0.444
Sand 0.290
Gold 0.129

Salt

Air

Lithium

Vegetable Oil
Chemistry
1 answer:
Firdavs [7]2 years ago
7 0

Answer:

LIthium vegetable oil

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jan is holding an ice cube. what causes the ice to melt? thermal energy from the ice is transferred to the air. thermal energy f
loris [4]

Answer: Ice is melting due to the transfer of thermal energy from Jan's hand to ice.

Explanation: The melting of ice is a physical change and is happening when the thermal energy from Jan's hand is transferred to ice. Due to this energy transfer, the particles of ice starts to move faster and hence, making the ice melt.

In this, the physical state of ice is changing from solid to liquid state.

H_2O(s)\rightleftharpoons H_2O(l)

8 0
2 years ago
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An unknown element X has the following isotopes: ⁵²X (89.00% abundant), ⁴⁹X (8.00% abundant), ⁵⁰X (3.00% abundant). What is the
Vlad [161]

Answer:

52 amu

Explanation:

To get the relative atomic mass of the element, we need to take into consideration, the atomic masses of the different isotopes and their relative abundances. We simply multiply the percentages with the masses. This can be obtained as follows:

[89/100 * 52] + [8/100 * 49] + [3/100 * 50]

46.28 + 3.92 + 1.5 =51.7 amu

The approximate atomic mass of element x is 52 amu

6 0
2 years ago
A transition in the balmer series for hydrogen has an observed wavelength of 434 nm. Use the Rydberg equation below to find the
DanielleElmas [232]

Answer:

i. n = 5

ii. ΔE = 7.61 × 10^{-46} KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × 10^{-18}(\frac{1}{n^{2}_{final} } - \frac{1}{n^{2}_{initial}  })

    (1/434 × 10^{-9}) = -2.178 × 10^{-18} (\frac{n^{2}_{initial} - n^{2}_{final}  }{n^{2}_{final} n^{2}_{initial}   })

⇒ 434 × 10^{-9} = (1/-2.178 × 10^{-18})\frac{n^{2}_{final} *n^{2}_{initial}   }{n^{2}_{initial} - n^{2}_{final}    }

But, n_{final} = 2

434 × 10^{-9} = (1/2.178 × 10^{-18})\frac{2^{2} n^{2}_{initial}  }{n^{2}_{initial} - 2^{2}  }

434 × 10^{-9}  × 2.178 × 10^{-18} = (\frac{4n^{2}_{initial}  }{n^{2}_{initial} - 4 })

⇒ n_{initial} = 5

Therefore, the initial energy level where transition occurred is from 5.

2. ΔE = hf

     = (hc) ÷ λ

    = (6.626 × 10−34 × 3.0 × 10^{8} ) ÷ (434 × 10^{-9})

    = (1.9878 × 10^{-25}) ÷ (434 × 10^{-9})

    = 4.58 × 10^{-19} J

    = 4.58 × 10^{-22} KJ

But 1 mole = 6.02×10^{23}, then;

energy in KJ/mole = (4.58 × 10^{-22} KJ) ÷ (6.02×10^{23})

         = 7.61 × 10^{-46} KJ/mole

7 0
2 years ago
Why a slow growing forest can have a very low NPP and yet store a massive amount of biomass.
drek231 [11]
Net Primary Productivity ... the amount of biomass present in an ecosystem at a particular time .... Explain why a slow growing forest can have a very low NPP and yet store a massive amount of biomass.
8 0
2 years ago
The metallic radius of a potassium atom is 231 pm. What is the volume of a potassium atom in cubic meters?
Sauron [17]
1 pm = 10∧-10 cm
Therefore, 230 pm is equivalent to 2.3 ×10∧-8 cm.
Atom is in the shape of a sphere,
The volume of a sphere is given by 4/3πr³
Thus, volume of the atom = 4/3π( 2.3 ×10∧-8)³
                                          = 4/3 (3.142 ×12.167×10∧-24
                                           = 5.096 ×10∧-23 cm³
but 1m³= 1000000cm³
Therefore, the volume of the atom = 5.096 ×10∧-29 m³
8 0
2 years ago
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