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2 years ago

Which layer in the diagram was the original layer?

Chemistry

2 answers:

tatiyna2 years ago

4 0

I’m almost sure it’s D but i could be wrong

Lynna [10]2 years ago

4 0

The correct answer is bedrock.

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In order to boil water, Jacy places a pan of water on the burner of a stove. By which process of thermal energy transfer does

nlexa [21]

Answer:

A campfire is a perfect example of the different kinds of heat transfer. If you boil water in a kettle, the heat is transferred through convection from the fire to the pot.

Explanation:

3 0

3 years ago

Uranium-238 (U-238) decays into lead-206 (Pb-206). Which of the following is a true statement?

noname [10]

The correct answer is D

8 0

3 years ago

How much energy is released by the decay of 3 grams of 20Th in the following reaction 230 Th - 226Ra + 'He (230 Th = 229.9837 g/

wolverine [178]

Answer : The energy released by the decay of 3 grams of 'Th' is $2.728\times 10^{-15}J$

Explanation :

First we have to calculate the mass defect $(\Delta m)$ .

The balanced reaction is,

$^{230}Th\rightarrow ^{226}Ra+^{4}He$

Mass defect = Sum of mass of product - sum of mass of reactants

$\Delta m=(\text{Mass of Ra}+\text{Mass of He})-(\text{Mass of Th})$

$\Delta m=(225.9771+4.008)-(229.9837)=1.4\times 10^{-3}amu=2.324\times 10^{-30}kg$

conversion used : $(1amu=1.66\times 10^{-27}kg)$

Now we have to calculate the energy released.

$Energy=\Delta m\times (c)^2$

$Energy=(2.324\times 10^{-30}kg)\times (3\times 10^8m/s)^2$

$Energy=2.0916\times 10^{-13}J$

The energy released is $2.0916\times 10^{-13}J$

Now we have to calculate the energy released by the decay of 3 grams of 'Th'.

As, 230 grams of Th release energy = $2.0916\times 10^{-13}J$

So, 3 grams of Th release energy = $\frac{3}{230}\times 2.0916\times 10^{-13}J=2.728\times 10^{-15}J$

Therefore, the energy released by the decay of 3 grams of 'Th' is $2.728\times 10^{-15}J$

5 0

3 years ago

Water has a boiling point of 100°C.

SashulF [63]

Answer: higher,because when salt is dissolved bonds form between the salt ions and the water molecules

5 0

3 years ago

Read 2 more answers

When 38.1 grams of a certain metal at a temperature of 90.0°C is added to 100.0mL of water at a temperature of 17.6°C in a perfe

MakcuM [25]

Answer:

a. qm = 627.3 J

b. qw = 627.3 J

c. C₂ = 227.4 J/kg.°C

Explanation:

Since, the calorimeter is completely insulated. Therefore,

Heat Lost by Metal = Heat Gained by water

qm = qw

qm = m₁C₁ΔT₁

where,

qm = heat lost by metal = ?

m₁ = mass of water = (density)(volume) = (1000 kg/m³)(100 mL)(10⁻⁶ m³/1 mL)

m₁ = 0.1 kg

C₁ = specific heat capacity of water = 4182 J/kg.°C

ΔT₁ = Change in Temperature of Water = 19.1°C - 17.6°C = 1.5°C

Therefore,

qm = (0.1 kg)(4182 J/kg.°C)(1.5°C)

qm = 627.3 J

Since,

qm = qw

qw = 627.3 J

qm = m₂C₂ΔT₂

where,

m₂ = mass of metal = 38.1 g = 0.0381 kg

C₂ = specific heat capacity of metal = ?

ΔT₂ = Change in Temperature of metal = 90°C - 17.6°C = 72.4°C

Therefore,

627.3 J = (0.0381 kg)(C₂)(72.4°C)

(627.3 J)/(0.0381 kg)(72.4°C) = C₂

C₂ = 227.4 J/kg.°C

6 0

3 years ago