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A sample of nickel is heated to 99.8 degrees C and placed in a coffee cup calorimeter containing 150.0g water at 23.5 degrees C.

vesna_86 [32]

Answer: The mass of nickel that was originally heated is 28.4 grams.

Explanation:

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

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

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

where,

q = heat absorbed or released

$m_1$ = mass of nickel = ?

$m_2$ = mass of water = 150.0 g

$T_{final}$ = final temperature = 25.0°C

$T_1$ = initial temperature of nickel = 99.8°C

$T_2$ = initial temperature of water = 23.5°C

$c_1$ = specific heat of nickel = 0.444 J/g°C

$c_2$ = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

$m_1\times 0.444\times (25.0-99.8)=-[150.0\times 4.186\times (25.0-23.5)]$

$m_1=28.4g$

Hence, the mass of nickel that was originally heated is 28.4 grams.

3 0

3 years ago

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An 8.42 g piece of metal with a specific heat of 1.020 J g-1 k-1 wasn't heated to an unknown temperature. The metal was then pla

ANTONII [103]

Answer:

$T = 61.344\,^{\textdegree}C$

Explanation:

The heat received by water is equal to the heat rejected by the piece of metal. That is to say:

$-Q_{piece} = Q_{water}$

$(8.42\,g) \cdot \left(1.020\,\frac{J}{g\cdot ^{\textdegree}C} \right) \cdot (T - 20.40\,^{\textdegree}C)= (44.2\,g)\cdot \left(4.187\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (20.40\,^{\textdegree}C-18.50\,^{\textdegree}C)$

$\left(8.588\,\frac{J}{^{\textdegree}C} \right)\cdot (T-20.40\,^{\textdegree}C) = 351.624\,J$

The initial temperature of the piece of metal is:

$T = 61.344\,^{\textdegree}C$

4 0

3 years ago

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Two aqueous sulfuric acid solutions containing 20.0 wt% H2SO4

kozerog [31]

Answer:

The feed ratio (liters 20% solution/liter 60% solution) is 3,08

Explanation:

In this problem you have a 20,0 wt% H₂SO₄ and a 60,0 wt% H₂SO₄ solutions.

100 kg of 20% solution are 100kg/1,139 kg/L = 87,8 L 

100kg×20wt% = 20 kg H₂SO₄. In moles:

20 kg H₂SO₄ × (1 kmol/98,08 kg) = 0,2039 kmol H₂SO₄≡ 203,9 mol

The final molarity 4,00M comes from:

$\frac{203,9 moles+ Xmoles}{87,8L + Yliters}$ (1)

Where X moles and Y liters comes from 60,0 wt% H₂SO₄

100 L of 60,0 wt% H₂SO₄ are:

100L× $\frac{1,498 kgsolution}{L}$ × $\frac{60 kg H_{2}SO_{4}}{100kgSolution}$ × $\frac{1kmol}{98,08kg H_{2}SO_{4}}$ = 0,9164 kmolH₂SO₄ ≡ 916,4 moles

That means:

X/Y = 916,4/100 = 9,164 (2)

Replacing (2) in (1):

Y(liters of 60,0 wt% H₂SO₄) = 28,52 L

Thus, feed ratio (liters 20% solution/liter 60% solution):

87,8L/28,52L = 3,08

I hope it helps!

8 0

3 years ago

What does the symbol ka mean in chemistry?

Dmitry_Shevchenko [17]

Ka is an equilibrium constant for the partial ionization of "weak" acids in water.

4 0

4 years ago

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How many atoms are present in 2.1 moles of Fe

Lyrx [107]

Answer:

1.3 x 10²⁴ atoms Fe

Explanation:

To convert moles to atoms, you need to multiply your given value by Avogadro's Number. This causes the conversion to occur because Avogadro's Number exists as a ratio. It is the ratio that allows for the cancellation of units during multiplication. This is why atoms should be in the numerator of your conversion. The final answer should have 2 sig figs to reflect the given value.

Avogadro's Number:

1 mole = 6.022 x 10²³ atoms

2.1 moles Fe 6.022 x 10²³ atoms
--------------------- x ------------------------------- = 1.3 x 10²⁴ atoms Fe
1 mole

4 0

2 years ago