What type of physical weathering is caused by the expansion of water?

What branch of science studies The Periodic Table?

Mrac [35]

Answers:

1. Chemistry

2. Cold

3. Heat

4. Protons and neutrons

5. Positive

6. Neutral

7. True

8. Nucleus

9. Protons

10. Nobel gasses

11. Dmitri Mendeleev

12. the bottom of the square

13. True

14. Helium

8 0

3 years ago

Balance the following Equation:

lisabon 2012 [21]

Answer:

HCl is the limiting reactant. It will completely be consumed (1.37 moles)

Option D is correct

Explanation:

Step 1: Data given

Mass of Zinc (Zn) = 50.0 grams

Mass of Hydrogen chloride (HCl) = 50.0 grams

atomic mass Zn = 65.38 g/mol

Molar mass HCl = 36.46 g/mol

Step 2: The balanced equation

Zn + 2HCl → ZnCl2 + H2

Step 3: Calculate moles

Moles = mass / molar mass

Moles Zn = 50.0 grams / 65.38 g/mol

Moles Zn = 0.764 moles

Moles HCl = 50.0 grams / 36.46 g/mol

Moles HCl = 1.37 moles

Step 4: Calculate limiting reactant

For 1 mol Zn we need 2 moles HCl to produce 1 mol ZnCl2 and 1 mol H2

HCl is the limiting reactant. It will completely be consumed (1.37 moles)

Zn is in excess. There will react 1.37/2 = 0.685 moles

There will remain 0.764 -0.685 = 0.079 moles

7 0

3 years ago

Calculate the energy released in each of the following fusion reactions. Give your answers in MeV. (a) 2H + 3H → 4He + n 17.54 C

Vikentia [17]

Answer:

a) E = 17.55 MeV

b) E = 18.99 MeV

c) E = 3.29 MeV

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

e) E = 4.075 MeV

Explanation:

Energy Released, $E = \triangle M * 931.5$

$\triangle M = \sum M_{product} - \sum M_{reactant}$

Mass of 1H, $M_{H} = 1.007823$

Mass of 2H, $M_{2H} = 2.0141u$

Mass of 3H, $M_{3H} = 3.016 u$

Mass of Helium, $M_{4He} = 4.002602u$

Mass of Beryllium, $M_{7Be} = 7.01693 u$

Mass of neutron, $M_{n} = 1.008664 u$

a) $2H + 3H \rightarrow 4He + n$

$\triangle M = (4M_{He} + M_{n} ) - (2M_{H} + 3 M_{H} )\\\triangle M = ( 4.0026 + 1.008664) - (2.0141 + 3.016 )\\\triangle M = -0.01884u$

Energy released,

$E = -0.01884 * 931.5\\E = -17.55 Mev$

Energy released = 17.55 MeV

b) $4He + 4He \rightarrow 7Be + n$

$\triangle M = (M_{7Be} + M_{n} ) - (M_{4He} + M_{4He} )\\\triangle M = ( 7.01693 + 1.008664) - (4.002602 + 4.002602 )\\\triangle M = 0.02039 u$

Energy released,

$E = 0.02039 * 931.5\\E = 18.99 Mev$

c) $2H + 2H \rightarrow 3 He$ + n

$\triangle M = (M_{3He} + M_{n} ) - (M_{2H} + M_{2H} )\\\triangle M = ( 3.016 + 1.008664) - (2.0141 + 2.0141 )\\\triangle M = -0.003536 u$

Energy released,

$E = -0.003536 * 931.5\\E = -3.29 Mev$

E = 3.29 MeV(Energy is released)

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

e) $2H + 2H \rightarrow 3H + 1H$

$\triangle M = (M_{3H} + M_{1H} ) - (M_{2H} + M_{2H} )\\\triangle M = ( 3.016 + 1.007825) - (2.0141 + 2.0141 )\\\triangle M = -0.00435 u$

$E = -0.00435 * 931.5\\E =-4.075 Mev$

E = 4.075 MeV ( Energy is released)

5 0

3 years ago

3) List two other names for P-waves and for S-waves.

Dimas [21]

3. Other names for S- waves are secondary waves, shear waves, and sometimes elastic S-waves. Other names for P-waves are primary waves and compressional waves.
4. You need 3 stations, because scientists find the difference between the arrival times of the primary and the secondary waves at each of the 3 stations, then the time difference is used to determine the distance of the epicentre from each station. The greater the difference in time, the further away the epicentre is. A circle is drawn around each station, with a radius corresponding to the epicentre’s distance from that station. The point where the three circles meet is the epicentre. If you only had two stations, you could only predict the epicentre, as the point where all three circles meet wouldn’t be complete, you’d have to try and estimate where the third one would intercept. This would greaten the chance of error and isn’t as accurate.

Hope this helps!

5 0

3 years ago

Calculate the heat of decomposition for this process at constant pressure and 25°C: CaCO3(s) → CaO(s) + CO2(g) The standard enth

ololo11 [35]

Answer:

177.8kJ/mol

Explanation:

In this reaction, the heat of decomposition is the same as the heat of formation. This is a decomposition reaction.

Given parameters:

ΔHf CaCO₃ = -1206.9kJ/mol

ΔHf CaO = −635.6 kJ/mol

ΔHf CO₂ = −393.5 kJ/mol

The heat of decomposition =

Sum of ΔHf of products - Sum of ΔHf of reactants

The equation of the reaction is shown below:

CaCO₃ → CaO + CO₂

The heat of decomposition = [ -635.6 + (-393.5)] - [−1206.9 ]

= -1029.1 + 1206.9

= 177.8kJ/mol

8 0

2 years ago