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SSSSS [86.1K]
3 years ago
14

How much heat, in joules and in calories, must be added to a 75.0–g iron block with a specific heat of 0.449 j/g °c to increase

its temperature from 25 °c to its melting temperature of 1535 °c?
Chemistry
1 answer:
Marina CMI [18]3 years ago
3 0
The amount of heat needed would be the specific heat multiplied by the mass of the substance and the temperature difference. In this case, the mass would be 75.0–g, the specific heat  would be 0.449 j/g °c, and the temperature difference would be <span>1535 -25= 1510

Then the calculation would be: </span>0.449 j/g °c * 75g * 1510°c = 50,849.25J

In calorie it would be: 50849.25J / 4.184J/cal= 12,153.26 calorie
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Calculate the number of sucrose molecules in a 75.0 gram sample.
zzz [600]

Answer:

1.32*10^23 molecules

Explanation:

sucrose formula: C12H22O11

molar mass: 12(12.01)+22(1.01)+11(16.00)=342.34g/mol

75.0 g C12H22O11 * (1 mol C12H22O11)/(342.34g C12H22O11)=0.219 mol C12H22O11

0.219 mol * (6.022*10^23)/mol = 1.32*10^23 molecules (three sig. figures)

5 0
3 years ago
All parts of the rock cycle are connected, whether directly or indirectly. Which statement
Ad libitum [116K]

Answer:

2. Igneous rocks can weather, creating sediments that form sedimentary rocks

Explanation:

Sedimentary rocks are formed from Igneous rocks when rocks are broken down by weathering.

5 0
3 years ago
For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the
Aneli [31]
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
8 0
4 years ago
What number of atoms of phosphorus are present in 1.00g of each of the compounds in exercise 40?
ira [324]

Answer:

Answer: 1.095 * 10^22 atoms of P.

Explanation:

3 0
3 years ago
A. Convert the mass of the Salt weighed out to fg.
omeli [17]

Answer:

6.564×10¹⁶ fg.

Explanation:

The following data were obtained from the question:

Mass of beaker = 76.9 g

Mass of beaker + salt = 142.54 g

Mass of salt in fg =?

Next, we shall determine the mass of the salt in grams (g). This can be obtained as follow:

Mass of beaker = 76.9 g

Mass of beaker + salt = 142.54 g

Mass of salt =?

Mass of salt = (Mass of beaker + salt) – (Mass of beaker)

Mass of salt = 142.54 – 76.9

Mass of salt = 65.64 g

Finally, we shall convert 65.64 g to femtograms (fg) as illustrated below:

Recall:

1 g = 1×10¹⁵ fg

Therefore,

65.64 g = 65.64 g × 1×10¹⁵ fg / 1g

65.64 g = 6.564×10¹⁶ fg

Therefore, the mass of the salt is 6.564×10¹⁶ fg.

3 0
3 years ago
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