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3 years ago

11

How many fluoride ion (F-1) are needed to balance a lead (IV) ion (Pb+4)?

2 answers:

mash [69]3 years ago

4 0

If your lead (IV) has a charge of +4 and the fluorine has a -1 then you would need 4 F-1 to balance the +4 charge.
PbF4

salantis [7]3 years ago

3 0

3

Pb2F3

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At which of the following temperatures does all molecular motion stop

Kamila [148]

All molecular motion stop at 0 k wich is zero kelvin. At absolute 0 it stops. The temperature of 0 entropy at which all molecular motion stops equals in centigrades to -273.15° C which is the same as 0 in kelvin degrees. Have in mind that t<span>emperature is a measure of the average kinetic energy of the </span>molecules<span> in a material.</span>

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3 years ago

Porfavor alguien que me explique los Cambios bioquimicos en los musculos ,sangre e higado durante los ejercicios fisicos y como

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3 years ago

. A reaction in which A, B, and C react to form products is first order in A, second order in B, and zero order in C. a) Write a

AleksAgata [21]

Answer:

a) r = k × [A] × [B]²

b) 3

Explanation:

Let's consider the following generic reaction

A + B + C ⇒ Products

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r = k × [A]ᵃ × [B]ᵇ × [C]ⁿ

where

r: reaction rate

k: rate constant

a, b, n: reaction orders

This reaction is first order in A, second order in B, and zero order in C. The rate law is:

r = k × [A]¹ × [B]² × [C]⁰

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3 years ago

Must show units and how they cancelli 1.) 175 km to um 3.) 385 nm to dm 5.) 492 um tom 7.) 52 x 103 dm to mm 9.) 321x 1035 mm to

morpeh [17]

Explanation:

1.) 175 km to μm

$1 km=10^9 \mu m$

$175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m$

3.) 385 nm to dm

$1 nm=10^{-8} dm$

$385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm$

5.) 492 μm to m

1 μm = $10^{-6} m$

$492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m$

7.) $52\times 10^3$ dm to mm

1 dm = 100 mm

$52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm$

9.) $321\times 10^{35}$ mm to km

$1 mm = 10^{-6} km$

$321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km$

11.) $456\times 10^3$ m to km

m = 0.001 km

$456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km$

13.) $422\times 10^3$ m to nm

$1 m = 10^{9} nm$

$422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm$

15.) $4.87\times 10^{30}$ m to pm

$1 m = 10^{12} pm$

$4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm$

17.) $5.26\times 10^3$ m to um

1 m = $10^{6} \mu m$

$5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m$

19.) $1.25\times 10^{35}$ m to Mm

1 m = $10^{-6} Mm$

$1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm$

21.) $4.22\times 10^3$ Tm to nm

$1 Tm = 10^{21} nm$

$4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm$

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3 years ago

What is the net Ionic equation of HClO4(aq)+ NaOH(aq) = H2O(l)+ NaClO4(aq)

uranmaximum [27]

H^+(aq) + OH^-(aq) ---> H2O(l)

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3 years ago