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kaheart [24]
2 years ago
13

In discharge tube gases are good conductors of electricity at very low pressure and high

Chemistry
1 answer:
Yuki888 [10]2 years ago
3 0

4)presence of ions and electrons

1)presence of atoms

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John is outstanding forward on his club select soccer team. He is know for his ball control and speed. After a game he often mak
natima [27]

Answer:

This self evaluation helps them to better his sport.

Explanation:

This is the right way to evaluate the performance of John because this sketch shows him the way he perform in the match. Due to this sketch he can better his performance and remove all the mistakes he had done in the match which results in better performance of John in the next match. This self evaluation helps them to better his sport and got new chances in order to select for a better club or national soccer team.

4 0
3 years ago
Draw an energy diagram for an endothermic and exothermic reaction and label the diagram ​
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Energy diagram for and endothermic and exothermic reaction

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3 years ago
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Do you think that humans are more likely to visit another
Darina [25.2K]

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terrestrial planets have a higher density than Jovian planets

7 0
3 years ago
A student performs an experiment to determine the density of a sugar solution. She obtains the following results: 1.79 g/mL, 1.8
ivann1987 [24]

Answer:

her results has good precision

Explanation:

accuracy and precision

4 0
3 years ago
A chemist dissolves 797. mg of pure barium hydroxide in enough water to make up 60. mL of solution. Calculate the pH of the solu
topjm [15]

Answer:

13.20

Explanation:

Step 1: Calculate the moles of Ba(OH)₂

The molar mass of Ba(OH)₂ is 171.34 g/mol.

0.797 g × 1 mol/171.34 g = 4.65 × 10⁻³ mol

Step 2: Calculate the molar concentration of Ba(OH)₂

Molarity is equal to the moles of solute divided by the liters of solution.

[Ba(OH)₂] = 4.65 × 10⁻³ mol/60 × 10⁻³ L = 0.078 M

Step 3: Calculate [OH⁻]

Ba(OH)₂ is a strong base according to the following equation.

Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻

The concentration of OH⁻ is 2/1 × 0.078 M = 0.16 M

Step 4: Calculate the pOH

pOH = -log OH⁻ = -log 0.16 = 0.80

Step 5: Calculate the pH

We will use the following expression.

pH + pOH = 14

pH = 14 - 0.80 = 13.20

7 0
3 years ago
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