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lions [1.4K]
2 years ago
14

Round off each of the following numbers to two significant figures:

Chemistry
2 answers:
Kay [80]2 years ago
7 0
A) 5.2 x 10^2
B) 86.
C) 6.4 x 10^3
D) 5.0
E) 22.
F) 0.89
strojnjashka [21]2 years ago
7 0

Answer:

(a) 5.1 × 10²

(b) 86

(c) 6.4 × 10³

(d) 5.0

(e) 22

(f) 0.89

Explanation:

To round off a number, we consider the number that is to its right. If that number is 5 or higher, we increase the number of interest. If the number to the right is 4 or lower, the number of interest remains the same. When taking less significant figures, we might need to use scientific notation for large or small numbers.

<em>Round off each of the following numbers to </em><em>two</em><em> significant figures: </em>

<em>(a) 517</em>   → 5.1 × 10²

<em>(b) 86.3</em>  → 86

<em>(c) 6.382 × 10³ </em> → 6.4 × 10³

<em>(d) 5.0008</em>  → 5.0

<em>(e) 22.497</em>  → 22

<em>(f) 0.885</em> → 0.89

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Answer:

Rotational spectroscopy, the dipole moment must change during the transition.

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Explanation:

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1 year ago
H2A and BOH are acid and base and they react according to the following balanced equation: H2A(aq) + 2 BOH(aq) → B2A(aq) + 2 H2O
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Answer:

ΔH=15000 J  =  15KJ

Explanation:

In this exercise  you have find the enthalpy of reaction this is the difference between enthalpy of reactans and products,

For the following equation

H2A(aq) + 2 BOH(aq) → B2A(aq) + 2 H2O(l)

We know that 0.20 moles of BOH reacted with excess amount of H2A solution and 1500. J

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for de reactions exothermics tha enthalpy is negative so:

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2 years ago
What is the value of ΔGo in kJ at 25 oC for the reaction between the pair: Mn(s) and Ag+(aq) to give Ag(s) and Mn2+(aq) Use the
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Answer:  -3.8\times 10^{5}J

Explanation:

Mn+2Ag^{+}\rightarrow Mn^{2+}+2Ag

Here Mn undergoes oxidation by loss of electrons, thus act as anode. silver undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

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E^0_{[Mn^{2+}/Mn]}= -1.18V

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E^0=E^0_{[Ag^{+}/Ag]}- E^0_{[Mn^{2+}/Mn]}

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\Delta G^0 = gibbs free energy

n= no of electrons gained or lost  = 2

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\Delta G^0=-2\times 96500\times (1.98)=-3.8\times 10^{5}J

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Nitrogen has a common valency of 3

The valency of each elements will determine the most likely value of x as outlined in the answer above.

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