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notka56 [123]
3 years ago
12

The atomic number of gold (Au) is 79, and it has a mass number of 197. How many electrons are present in each atom of gold?

Chemistry
1 answer:
aliya0001 [1]3 years ago
8 0
There would be 79 electrons present in each atom of gold. I hope this helps :)
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The shadow zone exists because ?
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At almost the opposite point on the Earth's surface, the "P" waves reappear. The shadow zone exists because the waves are refracted as they pass through the boundary between the mantle and the core and are diverted from their original paths.
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How do you know a physical change has occurred when evaporating water produces gaseous water?
levacccp [35]

Answer:

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6 0
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Which formula can be used to calculate the theoretical yield?
Bas_tet [7]

Answer:

Option C, (Actual yield ÷ percent yield) × 100

Explanation:

Theoretical yield is defined as the total amount of product formed for given reactants in a chemical reaction.  It is an ideal case which assumes no exceptions or wastage.

The mathematical relation between the actual yield, percent yield and theoretical yield is as follows -

P.Y. = \frac{M_{A.Y.}}{M_{T.Y.}} * 100

Where

P.Y. represents the percent yield a

M A.Y. represents the mass obtained from actual yield

M T.Y. represents the mass obtained from theoretical yield

Hence, if we rearrange the formula, we get -

M_{T.Y.} = \frac{M_{A.Y.}}{P.Y.} * 100

Hence, option C is correct

5 0
3 years ago
How many chlorine ions are required to bond with one aluminum ion
Monica [59]
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4 0
3 years ago
Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given
professor190 [17]

Answer:  -227 kJ

Explanation:

The balanced chemical reaction is,

C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]

-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]

\Delta H_{C_2H_2}=-227kJ

Therefore, the enthalpy change for C_2H_2 is -227 kJ.

7 0
3 years ago
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