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3 years ago

In nuclear fission, when the nucleus is hit the atom will ....what?

Chemistry

1 answer:

dimaraw [331]3 years ago

3 0

Answer:

When a large, fissile atomic nucleus such as uranium-235 or plutonium-239 absorbs a neutron, it may undergo nuclear fission. The nucleus splits into two or more lighter nuclei, releasing kinetic energy, gamma radiation, and free neutrons.

Explanation:

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I have the question<br> How many moles of CaO are produced as 0.248 moles of CaCO3 (s) is heated?

IRISSAK [1]

Moles are mass/Mr
or Concentrate × Volume in order to find the mole of CaO u need to find the mole of CO2 and take it away from CaCO3

7 0

3 years ago

Read 2 more answers

QUESTION 57 Select the diamagnetic ion. <br> A. Ni2+ B. Cu2+ C. Sr2+ D. Cr3+

VMariaS [17]

Answer:

Explanation:

5 0

3 years ago

The measurement

g100num [7]

Answer:

D. 25cm3

Explanation:

if you add a 3 at the end of a number it means volume, therefore making it a solid

3 0

3 years ago

A coffee-cup calorimeter contains 140.0 g of water at 25.1°C . A 124.0-g block of copper metal is heated to 100.4°C by putting i

Kisachek [45]

Answer:

(a) 3347 J; (b) 3043 J; (c) 58 J/K; (d) 35.5 °C

Explanation:

(a) Heat lost by copper

The formula for the heat lost or gained by a substance is

q =mCΔT

ΔT = T₂ - T₁= 30.3 °C - 100.4 °C = -70.1 °C = -70.1 K

q = 124.0 g × 0.385 J·K⁻¹g⁻¹ × (-70.1 K) = -3347 J

The negative sign shows that heat is lost.

The copper block has lost 3347 J.

(b) Heat gained by water

ΔT = 30.3 °C - 25.1 °C = 5.2 °C = 5.2 K

q = 140.0 g × 4.18 J·K⁻¹g⁻¹ × 5.2 K = 3043 J

The water has gained 3043 J.

(c) Heat capacity of calorimeter

Heat lost by Cu = heat gained by water + heat gained by calorimeter

The temperature change for the calorimeter is the same as that for the water.

ΔT = 5.2 K

$\begin{array}{rcl}\text{3347 J} & = & \text{3043 J} + C \times \text{5.2 K}\\\text{304 J} & = & 5.2C \text{ K}\\C & = & \dfrac{\text{304 J}}{\text{5.2 K}}\\\\& = & \text{58 J/K}\\\end{array}$

The heat capacity of the calorimeter is 58 J/K.

(d) Final temperature of water

$\begin{array}{rcl}\text{Heat lost by copper } + \text{Heat gained by water}& = &0 \\\text{Heat lost by copper}& = &-\text{Heat gained by water} \\m_{\text{Cu}}C_{\text{Cu}}\Delta T_{\text{Cu}}& = & -m_{\text{w}}C_{\text{w}}\Delta T_{\text{w}}\\\end{array}\\$

$\begin{array}{rcl}\text{124.0 g} \times \text{0.385 J$\cdot$K$^{-1}$g$^{-1}$}\times \Delta T_{\text{Cu}}& = & -\text{140.0 g} \times 4.18 \text{ J$\cdot$ K$^{-1}$g$^{-1}$}\times \Delta T_{\text{w}}\\\text{47.7 J$\cdot$K$^{-1}$}\times \Delta T_{\text{Cu}}& = &-\text{585 J$\cdot$ K$^{-1}$g}\times \Delta T_{\text{w}}\\\Delta T_{\text{Cu}} & = & -12.26\Delta T_{\text{w}}\\\end{array}$

$\begin{array}{rcl}\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26(\Delta T_{\text{f}} - 30.3\, ^{\circ}\text{C})\\\Delta T_{\text{f}} - 100.4 \, ^{\circ}\text{C} & = & -12.26\Delta T_{\text{f}} + 371\, ^{\circ}\text{C}\\13.26\Delta T_{\text{f}} & = & 471\, ^{\circ}\text{C}\\\Delta T_{\text{f}} & = & 35.5\, ^{\circ}\text{C}\\\end{array}$

The final temperature of the water would be 35.5 °C.

7 0

3 years ago

Considering the chart shown, at what temperature does the substance boil? 200ºC 1,200ºC 2,200ºC 2,400ºC

kodGreya [7K]

Answer:

The answer is 2,200ºC

Explanation:
I took the assignment for Edge, I don't think I can send the image because it might pick up on that and get reported, sorry!

The only other rationale that I have is that it's boiling because the graph shows that it's at a constant temperature/rate at 2,200ºC for quite a while. Typically when something boils, it stays at that constant rate of boiling, unless you turn the temperature up or it's finally able to peak..?

8 0

3 years ago