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Feliz [49]
3 years ago
12

Positive and Negative ions attract each-other and form chemical __________

Chemistry
1 answer:
e-lub [12.9K]3 years ago
7 0

Answer:

Changes

Explanation:

When positive and negative ions attract each other it fors a negative chemical changes

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A 5 gram round ball has a density of 1.25 grams/milliliter. What is the volume of the round ball?
Arlecino [84]

Explanation:

m=5g

density=1.25g/ml

density=m/v

v=m/density=5/1.25

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3 years ago
State Boyle's, Charles's, and Gay-Lussac's laws using sentences, then equations
dimaraw [331]

Explanation:

1. Boyle's Law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}      (At constant temperature and number of moles)

P_1\times {V_1}=P_2\times V_2

2. Charles' Law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

V\propto T    (At constant pressure and number of moles

\frac{V_1}{T_1}=\frac{V_2}{T_2}

3. Gay Lussac's Law states that tempertaure is directly proportional to the pressure of the gas at constant volume and number of moles of gas

P\propto T    (At constant volume and number of moles)

\frac{V_1}{n_1}=\frac{V_2}{n_2}

7 0
3 years ago
A sample of an ideal gas at 1.00 atm and a volume of 1.84 L was placed in a weighted balloon and dropped into the ocean. As the
Inessa05 [86]

Answer:

0.0613 L

Explanation:

Given data

  • Initial pressure (P₁): 1.00 atm
  • Initial volume (V₁): 1.84 L
  • Final pressure (P₂): 30.0 atm
  • Final volume (V₂): ?

Since we are dealing with an ideal gas, we can calculate the final volume using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁ / P₂

V₂ = 1.00 atm × 1.84 L / 30.0 atm

V₂ = 0.0613 L

6 0
4 years ago
a cube of iron (cp = 0.450 j/g•°c) with a mass of 55.8 g is heated from 25.0°c to 49.0°c. how much heat is required for this pro
krek1111 [17]
Q = ?

Cp = 0.450 j/g°C

Δt =  49.0ºC - 25ºC => 24ºC

m = 55.8 g

Q = m x Cp x Δt

Q = 55.8 x 0.450 x 24

Q = 602.64 J

hope this helps! 
7 0
3 years ago
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