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Gemiola [76]
4 years ago
13

Which state of matter is described as an electrically charged gas?

Chemistry
2 answers:
a_sh-v [17]4 years ago
4 0

Answer:

Plasma positive or negative charged gas

Explanation:

PLz brainliest

Ann [662]4 years ago
4 0

Answer:

It's plasma. it cant be anything else

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elena55 [62]
Turmeric solution turns red in contact with bases and is not affected by acids and neutral substances. Put one drop of each liquid as the turmeric paper. The liquid whose drop turns the paper red is solution hydroxide (base).
4 0
3 years ago
Helium decays to form lithium. Which equation correctly describes this decay?
GrogVix [38]

Answer:

⁴He₂ → ⁴Li₃ + ⁰e₋₁.

Explanation:

  • Helium decays with the release of an electron to form a lithium nucleus.
  • The number of nucleons is the same on both sides of the equation.
  • A neutron decays to form a proton with the release of an electron.
  • The atomic number increases from 2 to 3 while the atomic mass number stays the same at 4 nucleons.
8 0
3 years ago
Read 2 more answers
What is the ph of a peach with a [ –oh] = 3.2 × 10 –11 m?
3241004551 [841]

Answer:

Amerigo Vespucci, was the first European to reach the Caribbean Islands.

Please select the best answer from the choices provided

T

F

Explanation:

7 0
2 years ago
A 1.50-kilogram ball is attached to the end of a 0.520-meter string and swung in a circle. The velocity of the ball is 9.78 m/s.
Sedaia [141]

Answer:

F centripetal force (tension) = 275.9  N

Explanation:

Given data:

Mass = 1.50 kg

Radius = 0.520 m

Velocity of ball = 9.78 m/s

Tension = ?

Solution:

F centripetal force (tension) =  m.v² / R

F centripetal force (tension) = 1.50 kg . (9.78 m/s)² / 0.520 m

F centripetal force (tension) = 1.50 kg . 95.65 m²/s² / 0.520 m

F centripetal force (tension) = 143.5 kg. m²/s² / 0.520 m

F centripetal force (tension) = 275.9  N

7 0
3 years ago
Consider the reaction: CaCO3(s) à CaO(s) + CO2(g)
Vsevolod [243]

Answer:

131.5 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

First, we will calculate the standard enthalpy of the reaction (ΔH°).

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g) ) - 1 mol × ΔH°f(CaCO₃(s) )

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

ΔH° = 179.2 kJ

Then, we calculate the standard entropy of the reaction (ΔS°).

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g) ) - 1 mol × S°(CaCO₃(s) )

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

ΔS° = 160.2 J/K = 0.1602 kJ/K

Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.

ΔG° = ΔH° - T × ΔS°

ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K

ΔG° = 131.5 kJ

6 0
3 years ago
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