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3 years ago

5

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

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

so,

d=8.0g/4.0cm

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2 years ago

Which of the following is not an example of a chemical change?

elena55 [62]

This would be C. A car rusting

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3 years ago

Read 2 more answers

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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