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3 years ago

A piece of gold wire has a diameter of .175cm. How much will precisely 1.00 x 10^5 cm of the wire weigh

1 answer:

7 0

The main information we have to use here is the density of gold. From literature, the density of gold at room temperature is 19.32 g/cm³. To determine the mass, let's calculate the volume first. A wire is in the shape of a cylinder. Thus, the volume would be

V = πd²h/4
V = π(0.175 cm)²(1×10⁵ cm)/4
V = 2,405.28 cm³

Density = mass/volume
19.32 g/cm³ = Mass/2,405.28 cm³
Mass = 46,470 g gold wire

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How many joules of heat are required to heat 110 g of aluminum from 52.0 oC to 91.5 oC?

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Answer:

We need 3910.5 joules of energy

Explanation:

Step 1: Data given

Mass of aluminium = 110 grams

Initial temperature = 52.0 °C

Final temperature = 91.5 °C

Specific heat of aluminium = 0.900 J/g°C

Step 2: Calculate energy required

Q = m*c*ΔT

⇒with Q = the energy required = TO BE DETERMINED

⇒with m = the mass of aluminium = 110 grams

⇒with c = the specific heat of aluminium = 0.900 J/g°C

⇒with ΔT = the change in temperature = T2 - T1 = 91.5 °C - 52.0 °C = 39.5 °C

Q = 110 grams * 0.900 J/g°C * 39.5

Q = 3910.5 J

We need 3910.5 joules of energy

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3 years ago

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Answer:

the centers for disease control and prevention

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3 years ago

What’s the difference between 2o and o(subscript 2)?

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Answer:

2O is an atom while O2 Is a molecule

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3 years ago

Heat is transferred directly from a heat reservoir at 200 C to another heat reservoir at 5C. If the amount of heat transferred i

Daniel [21]

Answer:

ΔS=0.148 KJ/K

Explanation:

Given that

Q = 100 KJ

T₁=200°C

T₁=200+273 = 437 K

T₂=5°C

T₂=5 + 273 = 278 K

Reservoir 1 is rejecting heat that is why it taken as negative while the reservoir 2 is gaining the heat that is why it is taken as positive.

So the total change in entropy given as

ΔS= - Q/T₁ + Q/T₂

ΔS= - 100/473 + 100/278 KJ/K

ΔS=0.148 KJ/K

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3 years ago

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Answer:

CH4+2O2→2H2O+CO2

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3 years ago

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