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Vinil7 [7]
3 years ago
6

If you wanted to perform a controlled experiment to test the effect of temperature on the pressure of a bicycle tire, which of t

he following would be necessary?
Chemistry
1 answer:
Sav [38]3 years ago
6 0
It would be the needle pressure gage temp check
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How should scientist deal with the issue of endangered animals? write in complete 5 sentences please help
ANEK [815]
I think that scientist should build a habitat we’re the endangered animals could go free.By scientists building a habitat I mean giving the endangered animals a space to feel safe.Another way scientists could help endangered animals , would be to try and reproduce, the animal.For example scientist could help out with giving the animals hormones to be able to reproduce.And lastly scientists could do experimentation on the endangered animals to see if the animal ever were to go extinct how they could reproduce a new species.
3 0
3 years ago
Which contains elements with similar properties in the periodic table? A.) a period B.) a column C.) a row
zysi [14]

b............................................

3 0
3 years ago
Read 2 more answers
If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?
olga_2 [115]

Answer:

pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O   ⇄ H₃O⁺ + A⁻    with  Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

Equilibrium, M    0.40 - x                              x                               x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

4 0
3 years ago
What is exchanged between the system and the surroundings ​
BartSMP [9]

Answer:

see note under explanation

Explanation:

When describing system and surroundings the system is typically defined as the 'object of interest' being studied and surroundings 'everything else'. In thermodynamics heat flow is typically defined as endothermic or exothermic. However, one should realize that the terms endothermic and exothermic are in reference to the 'system' or object of interest being studied. For example if heat is transferred from a warm object to a cooler object it is imperative that the system be defined 1st. So, with that, assume the system is a warm metal cylinder being added into cooler water. When describing heat flow then the process is exothermic with respect to the metal cylinder (the system) but endothermic to the water and surroundings (everything else).

8 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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