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agasfer [191]
3 years ago
7

A hot metal plate at 150°C has been placed in air at room temperature. Which event would most likely take place over the next fe

w minutes?
-Molecules in both the metal and the surrounding air will start moving at lower speeds. -Molecules in both the metal and the surrounding air will start moving at higher speeds. -The air molecules that are surrounding the metal will slow down, and the molecules in the metal will speed up.
-The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Chemistry
1 answer:
aleksley [76]3 years ago
7 0

Answer:

until the next harvest, and seed must be held for the next season's ... successful grain storage is the moisture content of the crop. ... or both. If ambient temperatures are low, then air alone may cool the ... allow some of the drying to take place naturally while the crop ... employed to cool grain that has been placed in storage.

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Calculate the molecular (formula) mass of each compound: (a) iron(ll) acetate tetrahydrate; (b) sulfur tetrachloride; (c) potass
Nutka1998 [239]

Answer:

a) Iron(ll) acetate tetrahydrate: 245,68 g/mol

b) Sulfur tetrachloride: 173,87 g/mol

c) Potassium ermanganate 158,034 g/mol

Explanation:

To solve this kind of exercises you must look for the number of atoms in each molecule first, then look on the periodic table the atom weight and the multiply the atom weight times the quantity of each atom. For instance:

The molecule of Iron(II) acetate itetrahydrate is (CH3COO)2Fe•4H2O, it means that you have:

2 atoms of carbon times the atomic weight of C (12.00g/mol)= 24g

14 atoms of Hidrogen times the atomic weight of H (1,00g/mol)= 14g

6 atoms of Oxigen times the atomic weight of O (16,0g/mol)= 96

2 atoms of Iron times the atomic weight of Fe (55,84g/mol)= 111,68g

At last, you only have to add the results: 24+14+96+11,68= 245,68g/mol. This example was for the first molecule.

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5 0
3 years ago
The standard enthalpy of formation of BrCl(g) is 14.7 kJmol-1 . The standard enthalpies for the atomization of Br2(l) and Cl2(g)
natita [175]

Explanation:

Equation of the reaction:

Br2(l) + Cl2(g) --> 2BrCl(g)

The enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, BrCl.

The standard enthalpy change of formation for a compound,

ΔH°f, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of 1 atm.

This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction

1/2Br2(g) + 1/2Cl2(g) → BrCl(g)

Here, ΔH°rxn = ΔH°f

This means that the enthalpy change for this reaction will be twice the value of ΔH°f = 2 moles BrCl

Using Hess' law,

ΔH°f = total energy of reactant - total energy of product

= (1/2 * (+112) + 1/2 * (+121)) - 14.7

= 101.8 kJ/mol

ΔH°rxn = 101.8 kJ/mol.

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Answer:

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