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kherson [118]
3 years ago
5

How do matter cycles demonstrate the conservation of matter in a system?

Chemistry
1 answer:
damaskus [11]3 years ago
4 0

Answer:

These systems are defined both by the types of energy and matter they contain and by how that matter and energy move through and between systems. In natural systems, both energy and matter are conserved within a system. This means that energy and matter can change forms but cannot be created or destroyed.

Explanation:

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If a radioactive isotope of thorium (atomic number 90, mass number 232) emits 6 alpha particles and 4 beta particles during the
kolbaska11 [484]

Answer:

The mass number of the stable daughter product is 208

Explanation:

First thing's first, we have to write out the equation of the reaction. This is given as;

²³²₉₀Th → 6 ⁴₂α  +  4 ⁰₋₁ β + X

In order to obtain the identity of X, we have to obtain it's mass numbers and atomic number.

There is conservation of matter so we expect the mass number to remain the same in both the reactant and products.

Mass Number

Reactant = 232

Product = (6* 4 = 24) + (4 * 0 = 0) + x = 24 + x

since reactant = product

232 = 24 + x

x = 232 - 24 = 208

Atomic Number

Reactant = 90

Product = (6* 2 = 12) + (4 * -1 = -4) + x = 8 + x

since reactant = product

90 = 8 + x

x = 90 - 8 = 82

5 0
4 years ago
2. Jaimin takes Rs. 10000 at 12% per year as rate of interest, for 2 yean. How much interest he
Annette [7]

Answer:

2400

Explanation:

SI=PRT

p=10000

r=0.12

t=2

si=10000×0.12×2

=2400

3 0
3 years ago
what is the structure of XeO2F2???i know its hybridization as sp3d but cannot quite understand its diagram in free space,help an
skelet666 [1.2K]
The shape of XeO₂F₂ is Trigonal bi-pyramidal see-saw tetrahedron (see attached pictures)
- As you said the hybridization of Xe here is sp³d so its geometry has to be Trigonal bi-pyramidal in which F atom located on axial positions but for the final shape we exclude lone pair on Xe to give see-saw shape (see second picture) 
- Remember that we have 5 pairs (4 bond pairs + 1 lone pair) and we have to place lone pair at equatorial position.

6 0
3 years ago
Approximately how many ice cubes must melt to cool 650 milliliters of water from 29°C to 0°C? Assume that each ice cube contains
qwelly [4]

Answer : The number of ice cubes melt must be, 13

Explanation :

First we have to calculate the mass of water.

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}

Density of water = 1.00 g/mL

Volume of water = 650 mL

\text{Mass of water}=1.00g/mL\times 650mL=650g

Now we have to calculate the heat released on cooling.

Heat released on cooling = m\times c\times (T_2-T_1)

where,

m = mass of water = 650 g

c = specific heat capacity of water = 4.18J/g^oC

T_2 = final temperature = 29^oC

T_2 = initial temperature = 0^oC

Now put all the given values in the above expression, we get:

Heat released on cooling = 650g\times 4.18J/g^oC\times (29-0)^oC

Heat released on cooling = 78793 J = 78.793 kJ   (1 J = 0.001 kJ)

As, 1 ice cube contains 1 mole of water.

The heat required for 1 ice cube to melt = 6.02 kJ

Now we have to calculate the number of ice cubes melted.

Number of ice cubes melted = \frac{\text{Total heat}}{\text{Heat for 1 ice cube}}

Number of ice cubes melted = \frac{78.793kJ}{6.02kJ}

Number of ice cubes melted = 13.1 ≈ 13

Therefore, the number of ice cubes melt must be, 13

3 0
4 years ago
Helium decays to form lithium.
hjlf

Answer:

Natural nuclear disintegration.

Explanation:

This is because it cannot be managed, like it needs no energy particle yo occur, so its natural for helium to decay.

But it is nuclear because helium decays to release an energy particle (Beta particle).

7 0
3 years ago
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