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loris [4]
3 years ago
8

The strong nuclear force is always smaller then the electrostatic force true or false

Chemistry
2 answers:
AlekseyPX3 years ago
4 0

Answer: false

Explanation:

Gravitational, electrostatic, strong nuclear, and weak nuclear

Y_Kistochka [10]3 years ago
3 0

Answer: I think that the answer was False.

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What is the solute when stirring salt in water until the salt disappears?
Umnica [9.8K]

Answer:

The solute is the substance being dissolved.

The solvent is the substance dissolving the solute.

Therefore, the salt is the solute and the water is the solvent.

Explanation:

The salt is the solute.

3 0
2 years ago
Isaac the robot needs to travel from his charging station to school. Isaac travels 5 km south, 3 km west, 2 km north and 3 km ea
ladessa [460]

Answer:

13 km  

Explanation:

Distance travelled = 5 km + 3 km + 2 km + 3 km = 13 km

 

3 0
2 years ago
How much energy is needed to warm 7.40g of water by 55 degress C?
zloy xaker [14]
<span>E = mCdT
E = energy, m = mass, C = specific heat capacity, dT = change in temperature.
526 = 0.074C x 17
E = 0.074C x 55
Divide the equations
E/526 = (0.074C x 55)/(0.074C x 17) = 55/17
E = (55 x 526)/17 = 1702 J</span>
8 0
3 years ago
he heat of fusion of tetrahydrofuran is . Calculate the change in entropy when of tetrahydrofuran melts at . Be sure your answer
lana [24]

Answer:

\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}

Explanation:

Hello.

In this case, given the heat of fusion of THF to be 8.5 kJ/mol and freezing at -108.5 °C, for the required mass of 5.9 g, we can compute the entropy as:

\Delta S=\frac{n*\Delta H}{T}

Whereas n accounts for the moles which are computed below:

n=5.9g*\frac{1mol}{72g} =0.082mol

Thus, the entropy turns out:

\Delta S=\frac{0.0819mol*8.5 kJ/mol}{(-108.5+273.15)K}\\\\\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}

Best regards.

3 0
3 years ago
When sulfate (SO42-) serves as the electron acceptor at the end of a respiratory electron transport chain, the product is?
o-na [289]

When sulfate (SO₄²⁻) serves as the electron acceptor at the end of a respiratory electron transport chain, the product is hydrogen sulfide (H₂S).

How sulfate acts as electon acceptor and electron donor?

  • Sulfate (SO₄²⁻) is used as the electron acceptor in sulfate reduction, which results in the production of hydrogen sulfide (H2S) as a metabolic byproduct.
  • Many Gram negative bacteria identified in the -Proteobacteria use sulfate reduction, which is a rather energy-poor process.
  • Gram-positive organisms connected to Desulfotomaculum or the archaeon Archaeoglobus also utilise it.
  • Electron donors are needed for sulfate reduction, such as hydrogen gas or the carbon molecules lactate and pyruvate (organotrophic reducers) (lithotrophic reducers).

Learn more about the Electron transport chain with the help of the given link:

brainly.com/question/24372542

#SPJ4

3 0
2 years ago
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