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myrzilka [38]
4 years ago
6

Fusion reactors are more desirable than fission because:

Chemistry
2 answers:
Julli [10]4 years ago
5 0

Answer:

A, FUEL IS MORE READILY AVAILABLE

Serga [27]4 years ago
4 0
I believe it is A)fuel is more readily available:)
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A solution of 0.0027 M K2CrO4 was diluted from 3.00 mL to 100. mL. What is the molarity of the new solution?
Stells [14]

Answer:

[K₂CrO₄] → 8.1×10⁻⁵ M

Explanation:

First of all, you may know that if you dilute, molarity must decrease.

In the first solution we need to calculate the mmoles:

M = mmol/mL

mL . M = mmol

0.0027 mmol/mL . 3mL = 0.0081 mmoles

These mmoles  of potassium chromate are in 3 mL but, it stays in 100 mL too.

New molarity is:

0.0081 mmoles / 100mL = 8.1×10⁻⁵ M

4 0
3 years ago
Text me if im ur type im looking for a girlfreind
jasenka [17]

Answer:

Argon is an element on the periodic table. It is a member of the noble gasses, meaning that it is a monoatomic pure gas. In other words, argon gas particles are simply pure single atoms of argon. ... Finally, water is a polyatomic molecule H2O composed of two atoms of hydrogen and one atom of oxygen.

Explanation:

3 0
3 years ago
Read 2 more answers
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degree
lions [1.4K]

Answer:

The new temperature of the water bath 32.0°C.

Explanation:

Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)

Initial temperature of the water = T_1=33.9^oC=33.9+273K=306.9 K

Final temperature of the water = T_2

Specific heat capacity of water under these conditions =  c = 4.18 J/gK

Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J

( 1kJ=1000 J)

Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)

-69.0\times 1000 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)

-69,000.0 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)

T_2=304.86 K=304.86 -273^oC=31.86^oC\approx 32.0^oC

The new temperature of the water bath 32.0°C.

5 0
4 years ago
Predict whether the changes in enthalpy, entropy, and free energy will be positive or negative for the boiling of water, and exp
Sedbober [7]

Answer:

ΔH > 0; ΔS >0; ΔG = 0

Not spontaneous when T < 100 °C;

         Equilibrium when T = 100 °C

      Spontaneous when T > 100 °C

Step-by-step explanation:

The process is

H₂O(ℓ) ⇌ H₂O(g)

ΔH > 0 (positive), because we must <em>add heat</em> to boil water

ΔS > 0 (positive), because changing from a liquid to a gas i<em>ncreases the disorder </em>

ΔG = 0, because the liquid-vapour equilibrium process is at <em>equilibrium</em> at 100 °C

ΔG = ΔH – TΔS

Both ΔH and ΔS are positive.

If T = 100 °C, ΔG =0. ΔH = TΔS, and the system is at equilibrium.


If T < 100 °C, the ΔH term will predominate, because T has decreased below the equilibrium value.

ΔG > 0. The process is not spontaneous below 100 °C.


If T > 100 °C, the TΔS term will predominate, because T has increased above the equilibrium value.

ΔG < 0. The process is spontaneous above 100 °C.

4 0
4 years ago
I need the answer to this
Hatshy [7]

Answer:

#1 stationary

#2 warm

#3 cold.

Explanation:

hope this helps.

brainliest pls.

have a good day

7 0
2 years ago
Read 2 more answers
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