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Andrews [41]
3 years ago
7

Frank has a sample of steel that weighs 80 grams. If the density of his sample of steel is 8 g/cm3 ,what is the samples volume?

Chemistry
1 answer:
kirill115 [55]3 years ago
4 0

Answer: 10 cm3

Explanation:

Mass of sample of steel (m) = 80 grams

density of sample of steel (p) = 8 g/cm3 samples volume (v) = ?

Since density is obtained by dividing mass by volume, the density of the steel sample is expressed as:

density = mass / volume

p = m / v

make v the subject formula

v = m / p

v = 80 grams / 8 g/cm3

v = 10 cm3

Thus, the samples volume is 10 cm3

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Which two of Earth's spheres interact when the rainfall in a region increases after a volcanic eruption?
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Geosphere and Hydrosphere

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8 0
3 years ago
Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f Kf of 5.12 oC/m. With the a
kiruha [24]

From the calculation, the molar mass of the solution is 141 g/mol.

<h3>What is the molar mass?</h3>

We know that;

ΔT = K m i

K = the freezing constant

m = molality of the solution

i = the Van't Hoft factor

The molality of the solution is obtained from;

m = ΔT/K i

m = 3.89/5.12 * 1

m = 0.76 m

Now;

0.76 =  26.7 /MM/0.250

0.76 = 26.7 /0.250MM

0.76 * 0.250MM =  26.7

MM= 26.7/0.76 * 0.250

MM = 141 g/mol

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5 0
2 years ago
Suppose that 0.48 g of water at 25∘C∘C condenses on the surface of a 55-gg block of aluminum that is initially at 25∘C∘C. If the
GarryVolchara [31]

Answer : The final temperature of the metal block is, 25^oC

Explanation :

heat_{absorbed}=heat_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of aluminum = 55 g

m_2 = mass of water = 0.48 g

T_{final} = final temperature = ?

T_1 = temperature of aluminum = 25^oC

T_2 = temperature of water = 25^oC

c_1 = specific heat of aluminum = 0.900J/g^oC

c_2 = specific heat of water= 4.184J/g^oC

Now put all the given values in equation (1), we get

55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]

T_{final}=25^oC

Thus, the final temperature of the metal block is, 25^oC

6 0
3 years ago
Which has more thermal energy, 500g of ice or 500g of steam? explain your answer.
RUDIKE [14]
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