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3 years ago

Which of the following is not among the three likely sources of heat during the formation of the Earth?

Chemistry

1 answer:

Romashka-Z-Leto [24]3 years ago

5 0

Answer:

Constant volcanic eruption

Explanation:

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Air enters a diffuser witha velocity of 400 m/s, a pressure of 1 bar and temperature of 25 C. It exits with a temperature of 100

ANEK [815]

Answer:

Exit velocity of air is 96.43 m/s.

Explanation:

Given that

$V_1=400\ m/s$

$T_1=25C$

$T_2=100C$

For air

$C_p=1.005\ KJ/kg.K$

Now from first law of thermodynamics for open system at steady state

$h_1+\dfrac{V_1^2}{2000}+Q=h_2+\dfrac{V_2^2}{2000}+w$

For diffuser

$h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}$

We know that

$h=C_pT$

$h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2000}$

$1.005\times 25+\dfrac{400^2}{2000}=1.005\times 100+\dfrac{V_2^2}{2}$

$V_2=96.43\ m/s$

So the exit velocity of air is 96.43 m/s.

8 0

4 years ago

Complete the following analogy:

Jlenok [28]

The answer to this would be d. Precipitation patterns .

3 0

3 years ago

Read 2 more answers

A 2,31M solution of trans-4-methyl-2-pentene (C6H12, 85mL) is combined with 7,5mL of 3,55M elemental bromine to form an addition

kvasek [131]

Answer:

The statement is false

Explanation:

Number of moles of alkene = 2.31 × 85/1000 = 0.196 moles

Number of moles of Br2 = 3.55 × 7.5/1000 = 0.0266 moles

Given that the reaction is 1:1

1 mole of alkene reacts with 1 mole of bromine

0.196 moles of alkene should react with 0.196 moles of bromine

Hence, to achieve 100%yield, 0.196 moles of bromine and not 25mmols of elemental bromine

6 0

3 years ago

How many molecules of carbon dioxide are in 388.1 moles

crimeas [40]

Answer:

The answer is 53×11×10

7 0

3 years ago

a sample of cobalt (specific heat of Co=0.418J/G C )at 100.0 C is dropped into a calorimeter containing 500.0 mL of water at 21.

Alex17521 [72]

Answer:

$m_{Co}=6998g=7.0kg$

Explanation:

Hello there!

In this case, according to this equilibrium temperature problem, we can set up the following equation to relate the mass, specific heat and temperature change:

$Q_{Co}=-Q_{w}\\\\m_{Co}C_{Co}(T_f-T_{Co})=-m_{w}C_{w}(T_f-T_{w})$

Thus, we solve for the mass of cobalt as shown below:

$m_{Co}=\frac{-m_{w}C_{w}(T_f-T_{w})}{C_{Co}(T_f-T_{Co})} \\\\m_{Co}=\frac{-500.00g*4.184J/g\°C(67.1\°C-21.1\°C)}{0.418J/g\°C(67.1\°C-100\°C)} \\\\m_{Co}=6998g=7.0kg$

Best regards!

7 0

3 years ago