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2 years ago

11

In a post office, a 3-m long ramp is used to move carts onto a dock that is higher than 1 m. Which describes how the IMA of this

ramp compares with the IMA of the ramp that has a height of 1 m? It is greater. It depends on friction. It is the same. It is less.

1 answer:

pishuonlain [190]2 years ago

6 0

Answer:

well I think the answer is it depends on the friction

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You are presented with a white solid and told that due to careless labeling it is not clear if the substance is barium chloride,

Lady bird [3.3K]

The given salts are Barium chloride, lead chloride and zinc chloride.

The balanced chemical reaction between barium chloride and sodium sulfate:

$BaCl_{2}(aq)+Na_{2}SO_{4}(aq)-->2NaCl(aq)+BaSO_{4}(s)_{white precipitate}$

The balanced chemical reaction between lead chloride and sodium sulfate:

$PbCl_{2}(aq)+Na_{2}SO_{4}(aq)-->2NaCl(aq)+PbSO_{4}(s)_{white precipitate}$

The balanced chemical reaction between zinc chloride and sodium sulfate:

$ZnCl_{2}(aq)+Na_{2}SO_{4}(aq)-->2NaCl(aq)+ZnSO_{4}(aq)$

Both Barium chloride and lead chloride produce a white precipitate of their respective sulfates upon reaction with sodium sulfate. But BaCl2 is readily soluble in water where as Lead chloride is insoluble in water.

Therefore, the compound is barium chloride.

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1 year ago

When aqueous solutions of AgNO3 and KI are mixed, AgI precipitates. The balanced net ionic equation is ________. When aqueous so

MakcuM [25]

Answer:

Ag⁺ (aq) + I¯ (aq) —> AgI (s)

Explanation:

We'll begin by writing the dissociation equation for aqueous AgNO₃ and KI.

Aqueous AgNO₃ and KI will dissociate in solution as follow:

AgNO₃ (aq) —> Ag⁺(aq) + NO₃¯ (aq)

KI (aq) —> K⁺(aq) + I¯(aq)

Aqueous AgNO₃ and KI will react as follow:

AgNO₃ (aq) + KI (aq) —>

Ag⁺(aq) + NO₃¯ (aq) + K⁺ (aq) + I¯(aq) —> AgI (s) + K⁺ (aq) + NO₃¯ (aq)

Cancel out the spectator ions (i.e ions that appears on both sides of the equation) to obtain the net ionic equation. The spectator ions are K⁺ and NO₃¯.

Thus, the net ionic equation is:

Ag⁺ (aq) + I¯ (aq) —> AgI (s)

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2 years ago

Dead space is the portion of the respiratory system that: Select one: A.. must be filled with air before gas exchange can take p

ololo11 [35]

Answer: Option B

Explanation:

Dead space is the volume of sir in the body which is inhaled but it does not takes part in the gaseous exchange this is because it does not contains alveoli.

Not all the air that is inhaled inside the body is used in the process of respiration.

The dead space in the lungs is the region which helps in conducting airways, the air is not perfused to the alveoli.

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2 years ago

The rate of disappearance of HCl was measured for the following reaction:

AURORKA [14]

Answer: (a) 0.000083M/s, 0.000069M/s, 0.000052M/s, 0.000034M/s

(b) average reaction rate between t=0.0min to t=430.0min =0.000049M/s

(c) The average rate between t=54.0 and t=215.0min is greater than the average rate between t=107.0 and t=430.0min

Explanation:

Average reaction rate = change in concentration / time taken

(a) <em>after 54mins, t = 54*60s = 3240s</em>

average reaction rate = (1.58 - 1.85)M / (3240 * 0.0)s

= -0.27M/3240

= 0.000083M/s

<em>after 107mins, t = 107*60s = 6420s</em>

average reaction rate = (1.36 - 1.58)M/ (6420 - 3240)s

= -0.22M/3180s

= 0.000069M/s

<em>after 215mins, t = 215*60s = 12900s</em>

average reaction rate = (1.02 - 1.36)M/ (12900 - 6420)s

= -0.34M/6480s

= 0.000052M/s

<em>after 430mins,t = 430*60 = 25800s</em>

average reaction rate = (0.580 - 1.02)M / (25800 - 12900)s

= -0.44M/12900s

= 0.000034M/s

(b) <em>average reaction rate between t=0.0min to t=430.0min</em>

= (0.580 - 1.85)M/ (25800 - 0.0)s

= -1.27M/25800s

=0.000049M/s

(c) average reaction rate between t = 54.0min and t = 215.0min

= (1.02 - 1.58)M / (12900 - 3240)s

= -0.56M/9660s

= 0.000058M/s

average reaction rate between t=107.0 and t=430.0min

= (0.580 - 1.36)M / (25800 - 6420)s

= 0.78M /19380s

= 0.000040M/s

Therefore the average rate between t=54.0 and t=215.0min is greater than the average rate between t=107.0 and t=430.0min

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2 years ago

2. A quantity of 1.922g of methanol (CH3OH) was burned in a constant-volume

Cerrena [4.2K]

Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:

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1 year ago