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mamaluj [8]
2 years ago
13

Marcus measured the masses and volumes of samples of four different substances, and he calculated their densities. The table sho

ws Marcus’s measured and calculated values. Substance Mass (g) Volume (cm3) Density (g/cm3) aluminum 5.7 2.1 2.7 copper 14.4 1.6 9.0 iron 9.5 1.2 7.9 titanium 8.6 1.8 4.8 Next, Marcus obtained another sample, which is made of one of the four substances that he had already measured. The table shows Marcus’s measured values for this unknown sample. Substance Mass (g) Volume (cm3) ? 9.5 2.1 What is the unknown sample made of?
Chemistry
1 answer:
Zina [86]2 years ago
4 0

<u>Given:</u>

Calculated density values-

Aluminum = 2.7 g/cm3

Copper = 9.0 g/cm3

Iron = 7.9 g/cm3

Titanium = 4.8 g/cm3

Unknown sample mass = 9.5 g

Sample volume = 2.1 cm3

<u>To determine:</u>

The identity of the unknown sample

<u>Explanation:</u>

'Density' is a physical parameter which can be used to identify the nature of the unknown substance.

Density = Mass/Volume

For the unknown sample

Density = 9.5 g/2.1 cm3 = 4.52 g/cm3

This matches closely with the calculated density of titanium

Ans: The unknown substance is made of titanium

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The radioactive atom R 88 210 a is an alpha emitter. What nucleus does it produce?
olasank [31]

Answer:

X 86 206

Explanation:

Radioactive atoms are nuclei that can under go disintegration to emit either an alpha particle, beta particle or gamma radiation. The process could be spontaneous or stimulated.

When a radioactive atom R 88 210 emits alpha particle, it would produce an element with atomic number 86 and mass number 206 i.e X 86 206. An alpha particle is usually a helium nucleus.

                     R^{210} ⇒ x^{206} + He^{4} + energy

5 0
3 years ago
What is the mass of 7 x 10^28 atoms of Fe?
Nostrana [21]

Answer:

6 x 10⁶ g Fe

Explanation:

Step 1: Set up dimensional analysis

7 x 10²⁸ atoms Fe (1 mol Fe/6.02 x 10²³ atoms Fe)(55.85 g Fe/1 mol Fe)

Step 2: Multiply, divide, and cancel out units

atoms Fe and atoms Fe cancel out.

mol Fe and mol Fe cancel out.

We should be left with g Fe.

7 x 10²⁸/6.02 x 10²³ = 116279 mol Fe

116279(55.85) = 6.49 x 10⁶ g Fe

Step 3: Sig figs

There is only 1 sig fig in this problem.

6.49 x 10⁶ g Fe ≈ 6 x 10⁶ g Fe

4 0
3 years ago
A 31.1 g wafer of pure gold, initially at 69.3 _c, is submerged into 64.2 g of water at 27.8 _c in an insulated container. what
KIM [24]
Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water 

Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.

From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water

At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
      = (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)- 
      = 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
       = (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
       = 268.356(T - 27.8)

Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C

Answer: 28.4 °C

3 0
3 years ago
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Explanation:

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3 years ago
Write an equation between the carbonate and the acid​
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Answer:

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