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attashe74 [19]
3 years ago
5

As you enter the lab, you find two bottles labeled "concentrated ammonium phosphate." In one paragraph, using your own words, de

scribe the steps you could take to change one of the bottles into a diluted solution, and one of the bottles into a saturated solution.
Physics
1 answer:
yaroslaw [1]3 years ago
7 0

Answer:

Butane is a four member carbon chain.

The properties of butane are as follows.

Butane is less denser than water.

In the presence of air, it readily burns to form water vapour and carbon dioxide. The reaction is as follows.

           

It has a boiling point of .

Butane has faint petroleum like odor.

Butane does not react with water.

Butane has weak dispersion force and water has strong hydrogen bonding.  The weak dispersion force is unable to break the hydrogen bonding hence, butane does not dissolve in water.

Thus, we can conclude that the chemical properties butane burns readily in air and does not react with water are true out of the given list of options.

Explanation:

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An object has a kinetic energy ke, and potential energy pe. it also has a rest energy e0. what is the object's total energy e?
Dmitriy789 [7]
Look that one up in you text book PG:678 that is if you got the same book as my friend<span />
7 0
3 years ago
The atomic mass of gold is 0.197 kg/mole. how many moles are in 0.566 kg of gold.
iogann1982 [59]

Explanation:

0.566kg *(1mol/0.197 kg)= 2.87 mol gold

note how the units cancel out, if the units do not cancel out (kg/kg=1) then u did something wrong

7 0
3 years ago
Read 2 more answers
The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres
12345 [234]

Answer:0.0704 kg

Explanation:

Given

initial Absolute pressure(P_1)=210+101.325=311.325

T_1=25^{\circ}\approx 298 K

V=0.025 m^3

T_2=50^{\circ}\approx 323 K

as the volume remains constant therefore

\frac{P_1}{T_1}=\frac{P_2}{T_2}

\frac{311.325}{298}=\frac{P_2}{323}

P_2=337.44 KPa

therefore Gauge pressure is 337.44-101.325=236.117 KPa

Initial mass m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}

m_1=0.91 kg

Final mass m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}

m_2=0.839

Therefore m_1-m_2=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back

4 0
3 years ago
An object accelerates from rest and travels 53 m west in 5.2 s. Determine the acceleration
zmey [24]

Answer:

20.4m/s²

Explanation:

Given parameters:

Initial velocity  = 0m/s

Distance  = 53m

Time  = 5.2s

Unknown:

Acceleration  = ?

Solution:

This is a linear motion and we use the right motion equation;

        S = ut  + \frac{1}{2}at²

S is the distance

u is the initial velocity

a is the acceleration

t is the time

 Insert the parameters and solve;

       53  = (0x 5.2) +  \frac{1}{2} x a x 5.2

       53  = 2.6a

         a = \frac{53}{2.6}  = 20.4m/s²

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3 years ago
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