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3 years ago

How Many Significant Figures are there in 505

Chemistry

1 answer:

cupoosta [38]3 years ago

7 0

Answer:

there are three significant digits.

Explanation:

Thus, 21.8, 0.283 and 567 all have three significant digits. 2. Zeros appearing between non-zero digits are significant. Thus, 505 and 0.206 have three significant digits, while 50,005 has five significant digits. (from google)

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Given the reaction that occurs in an electrochemical cell:

KatRina [158]

Answer:

c) +2 to 0

Explanation:

SO4 has a charge of -2, so the Cu attached to that has to be a +2 since the polyatomic molecule has no overall charge

Cu(s) is a solid metal and they have no charge, therefore it is zero

Copper undergoes Oxidation (gain of electrons)

8 0

2 years ago

Read 2 more answers

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Andru [333]

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2 years ago

How to know how many valence electrons are in an element

Alenkinab [10]

Answer:

Hope it helped

Explanation:

For neutral atoms, the number of valence electrons is equal to the atom's main group number. The main group number for an element can be found from its column on the periodic table. For example, carbon is in group 4 and has 4 valence electrons. Oxygen is in group 6 and has 6 valence electrons.

7 0

3 years ago

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

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

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

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

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

2 years ago

Read 2 more answers

In some reactions, Be behaves like a typical alkaline earth metal; in others, it does not. Complete and balance the following: (

mina [271]

The correct answer is BeCl_2(l)+2Cl^-(solvated)→BeCl_4^2-.

Evaluating be behavior to see :

how it differs from the other Group 2A (2) members.

In this reaction Be behaves like other alkaline earth metals

The complete equation can be given as

BeCl_2(l)+2Cl^-(solvated)→BeCl_4^2-

BeCl_2 tends to form a chloro bridged dimer in the vapour state, however at high temperatures of the order of 1200K, this dimer dissociates into the linear monomer.

BeCl_2 has a chain structure in its solid form. Each Be atom in this structure is surrounded by chlorine atoms, two of which are connected by conversion bonds and the remaining two by covalent coordinate connections. This chain structure is displayed.

To know more about BeCl₂(I) + Cl⁻ refer the link:

brainly.com/question/5017059

#SPJ4

4 0

1 year ago