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3 years ago

Dominic made the table below to organize his notes about mixtures,

Chemistry

1 answer:

Troyanec [42]3 years ago

6 0

Answer: Although it is possible to have more than one state, it is also possible to have only one state.

Explanation: Just took the test

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If you collect 1.75 L of hydrogen gas during a lab experiment, when the room temperature is 23ºC and the barometric pressure is

Marianna [84]

Answer:

$n=0.0747mol$

Explanation:

Hello,

In this case, since we can consider hydrogen gas as an ideal gas, we check the volume-pressure-temperature-mole relationship by using the ideal gas equation:

$PV=nRT$

Whereas we are asked to compute the moles given the temperature in Kelvins, thr pressure in atm and volume in L as shown below:

$n=\frac{105kPa*\frac{0.009869atm}{1kPa}*1.75L}{0.082\frac{atm*L}{mol*K}*(23+273.15)K} \\\\n=0.0747mol$

Best regards.

6 0

3 years ago

Determine the number of protons and neutrons in plutonium-239 and enter its symbol in the form azx.

Komok [63]

Answer:

number of protons: 94
number of neutrons: 145

239
       Pu
   94

Explanation:

1) The atomic number, Z, of plutonium is Z = 94, i.e plutonium 94 protons

2) Plutoniun-239 is the isotope with mass number 239.

3) Mass number = number of protons + number of neutrons =>

number of neutrons = mass number - number of protons = 239 - 94 = 145

4) The notation requires that you indicate the symbol of the element with the atomic number (number of protons) and the mass number.

You put the mass number as a superscript at the left side of the symbol and the atomic number as subscript to the left of the symbol.

So, in this case the symbol is Pu, the superscript to the left is 239 and the subscript to the left is 94.

239
       Pu
   94

7 0

3 years ago

The dirigible Hindenburg had 3.7E6 m3 of hydrogen in its gas bags at 1.1 atm and 7°C. What was the weight of the hydrogen in pou

zepelin [54]

Assuming that the gas is ideal, we can use the ideal gas equation PV=nRT to calculate for the number of moles. Then, multiply the molar mass of the gas to obtain the mass. We do as follows:

PV = nRT
n = PV / RT
n = 1.1 atm (3.7x10^9 L) / 0.08205 L-atm/mol-K (280.15) = 177061931.3 mol H2

Mass = (177061931.3 mol H2) 18.02 g/mol ( 1 kg / 1000g) ( 2.2 lb / 1 kg ) = 7019443.21 lb H2

4 0

3 years ago

A 27.5 −g aluminum block is warmed to 65.9 ∘C and plunged into an insulated beaker containing 55.5 g water initially at 22.1 ∘C.

Wittaler [7]

Answer:

The final temperature of aluminium ≈ 26.32 °C

Explanation:

Step 1: explain the problem

A 27.5 −g aluminum block is warmed to 65.9 ∘C and plunged into an insulated beaker containing 55.5 g water initially at 22.1 ∘C.

Step 2: Data given

We will use the formule : Q = mcΔT

with Q = heat transfer ( J)

with m = mass of the substance (g)

with c = specific heat ( J/g °C)

with ΔT = change in temperature ( in °C or K)

mass of aluminium = 27.5g

mass of water = 55.5g

specific heat of aluminium = 0.900J/g °C

specific heat of water = 4.186 J/g °C

initial temperature of aluminium T1= 65.9 °C

initial temperature of water T1 = 22.1 °C

final temperature of water and aluminium = TO BE DETERMINED

Step 3: Calculate the initial temperature

To find the final temperature, we have to use the following formule:

-(Mass of aluminium) * (caluminium)*(ΔT)) = (Mass of water) *(cwater)*(ΔT)

-27.5g (0.900)(T2 - 65.9) = 55.5g (4.184j/g °C) (T2- 22.1)

-24.75*(T2-65.9) = 232.212 *(T2-22.1)

-24.75T2 + 1631.025 = 232.212T2 -5131,8852

-256.962 T2 = -6762.9102

T2 = 26.32 °C

The final temperature of aluminium ≈ 26.32 °C

6 0

3 years ago

Why did the author include details about the cooking show he saw?

Stella [2.4K]

Answer:

he wants to explain why he started cooking

6 0

3 years ago

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