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saw5 [17]
3 years ago
7

A second unknown substance has a density of 80 g/cm and a mass of 570g. What is the volume?

Chemistry
1 answer:
Salsk061 [2.6K]3 years ago
6 0

Answer:

25 g

10 cm

3

=

2.5 g/cm

3

Explanation:

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In an isothermal gas chromatography experiment using an ECD detector, 1.69 nmols of nchlorohexane, C6H13Cl, was added as an inte
prohojiy [21]

Answer:

The amount of Chlorodecane in the unknown is 0.105nmols

Explanation:

a) Since the GC is in an isothermal state, Chlorohexane C6H13Cl (1.69 nmols) because of its lower boiling point will elute first and Chlorodecane C12H21Cl will elute second.

The area of the first peak corresponding to Chlorohexane is 32434 units.

The area of the second peak corresponding to chlorodecane is 2022 units.

Since the response factor of the compound is not given in question and considering the response factor is same for both the compounds, the answer will be as follow:

1.69 nmols of Chlorohexane gives 32434 units

How much of chlorodecane gives 2022 units

By cross multiplication;

Moles of Chlorodecane = 2022*1.69/32434

                                         =0.105nmols

4 0
3 years ago
Please help me with this.
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Answer A: Connect a wire coil to an ammeter. Move a bar magnet into and out of the wire coil as you observe the ammeter.

6 0
2 years ago
How many molecules of sodium nitrate are produced when 40 g of sodium azide, NaN3, react with oxygen?
quester [9]
NaN 3 + AgNO 3 AgN 3 + NaNO 3
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2 years ago
For many purposes we can treat ammonia NH3 as an ideal gas at temperatures above its boiling point of −33.°C. Suppose the temper
Keith_Richards [23]

Answer:

The new pressure will be 0.225 kPa.

Explanation:

Applying combined gas law:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1\text{ and }V_1 are initial pressure and volume at initial temperature T_1.

P_2\text{ and }V_2 are final pressure and volume at initial temperature T_2.

We are given:

P_1=0.29 kPa\\V_1=V\\P_2=?\\V_2=V+50\% of V=1.5 V

T_1=-25^oC=248.15 K

T _2 = 16^oC=289.15 K

Putting values in above equation, we get:

\frac{0.29 kPa\times V}{248.15 K}=\frac{P_2\times 1.5V}{289.15 K}

P_1=0.225 kPa

Hence, the new pressure will be 0.225  kPa.

4 0
3 years ago
The flow of energy through living systems can be modeled as food chains, food webs, or energy pyramids. The energy organisms nee
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Answer:

A.......

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