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pashok25 [27]
3 years ago
5

2. How many moles of salt are present in 1.5L of a 5.OM salt water solution?

Chemistry
1 answer:
madam [21]3 years ago
8 0

Answer:

There are 7.5 moles of salt

Explanation:

5.0M means that in every liter of solution, there are 5 moles of salt. So, 1.5L of solution times 5 moles per liter equals 7.5 moles

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As part of a science experiment, Jose did a test for starch on a slice of apple and a slice of potato. The yellow-orange iodine
juin [17]

Answer:

See explanation

Explanation:

Iodine solution is used to test for starch.  A positive test for starch gives a blue-black color.

The fact that the color of the apple remained the same is indicative of the fact that starch was not contained in the apple.

A change in the color of potato indicates the presence of starch in the potato.

The fact that iodine did not react with apple should not be taken to mean that apples contain no starch at all. Starch changes gradually to sugar as fruits ripen. This is why the apple gave a negative test for starch.

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Find the density of metal with a volume 4.0 cm and a mass of 8.0 grams.
olga_2 [115]
D=m/v

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2 years ago
Which of the following is not an example of a chemical change?
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Starting with 2.50 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0C, a chemist first heats the gas at con
arsen [322]

Answer:

a)  T_b=590.775k

b)  W_t=1.08*10^4J

d)  Q=3.778*10^4J

d)  \triangle V=4.058*10^4J

Explanation:

From the question we are told that:

Moles of N2 n=2.50

Atmospheric pressure P=100atm

Temperature t=20 \textdegree C

                      t = 20+273

                     t = 293k

Initial heat Q=1.36 * 10^4 J

a)

Generally the equation for change in temperature is mathematically given by

\triangle T=\frac{Q}{N*C_v}

  Where

  C_v=Heat\ Capacity \approx 20.76 J/mol/K

T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }

T_b-293k=297.775

T_b=590.775k

b)

Generally the equation for ideal gas is mathematically given by

 PV=nRT

For v double

 T_c=2*590.775k

 T_c=1181.55k

Therefore

PV=Wbc

Wbc=(2.20)(8.314)(1181_590.778)

Wbc=10805.7J

Total Work-done W_t

W_t=Wab+Wbc

W_t=0+1.08*10^4

W_t=1.08*10^4J

c)

Generally the equation for amount of heat added is mathematically given by

Q=nC_p\triangle T

Q=2.20*2907*(1181.55-590.775)\\

Q=3.778*10^4J

d)

Generally the equation for change in internal energy of the gas is mathematically given by

\triangle V=nC_v \triangle T

\triangle V=2.20*20.76*(1181.55-293)k

\triangle V=4.058*10^4J

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