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d1i1m1o1n [39]
3 years ago
5

Under certain conditions, neon (Ne) gas diffuses at a rate of 7.0 centimeters per second. Under the same conditions, an unknown

gas diffuses at a rate of 4.9 centimeters per second. What is the approximate molar mass of the unknown gas?
Chemistry
1 answer:
yanalaym [24]3 years ago
3 0
A Scottish scientist Graham, Thomas formulated Graham's law of effusion or simply Graham's law. This law states that rate of gas 1 over gas 2 is equal to the square root of the molar mass of gas 2 over <span>he square root of the molar mass of gas 1. Since the molar mass of Neon is 4, the answer is then 8.16 g/mol</span>


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N2(g) + O2 (g) = 2 NO (g)

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4 years ago
What is the specific heat capacity of an unknown metal if 55.00 g of the metal initially at 90°C is placed in 75mL of water with
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Answer:

The specific heat capacity of the unknown metal is <u>1.04 J/g°C.</u>

Explanation:

Let the specific heat capacity of the unknown metal be x.

Given:

Mass of the metal is, m=55\ g

Initial temperature of the metal is, T_i=90\ \°C

Volume of water is, V=75\ ml

Specific heat capacity of water is, c_w=4.186\ J/g\°C

Initial temperature of water is, T_{wi}=25\ \°C

Final temperature of the system is, T=35\ \°C

We know that density of water is equal to 1 g/ml.

Mass is given as the product of density and volume.

Therefore, mass of water is given as:

m_w=1\times 75=75\ g

Now, fall of temperature of the unknown metal is given as:

\Delta T_m=T_i-T=90-35=55\ \°C

Rise of temperature of water is given as:

\Delta T_w=T-T_{wi}=35-25=10\ \°C

Now, as per conservation of energy,

Heat lost by metal = Heat gained by water

⇒ mx\Delta T_m=m_wc_w\Delta T_w

Plug in all the given values and solve for x. This gives,

55\times x\times 55=75\times 4.186\times 10\\3025x=3139.5\\x=\frac{3139.5}{3025}=1.04\ J/g\°C

Therefore, the specific heat capacity of the unknown metal is 1.04 J/g°C.

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