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gtnhenbr [62]
2 years ago
11

What is the frequency of a light that has a wavelength of 3.0 x 10 -7m?

Chemistry
1 answer:
alexdok [17]2 years ago
7 0

Answer:

 frequency of light (f) = 1 x 10¹⁵s⁻¹

Explanation:

Given Data:

Wavelength of light λ = 3.0 x10⁻⁷m

Frequency of light: to be calculated

Formula Used to find frequency:

                                  f = V/λ ........................... (1)

where

f is the frequency

V is the velocity

λ is wavelength

Velocity of light = 3 x 10⁸ ms⁻¹

put the values in equation (1)

                                     f =  3 x 10⁸ ms⁻¹ / 3.0 x10⁻⁷m

                                     f = 1 x 10¹⁵s⁻¹

   So the frequency of light =  1 x 10¹⁵s⁻¹

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What do all ions have in common in terms of their electrical structure?????
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Answer:

The ratio between protons to electrons is not 1:1

Explanation:

A normal atom will be neutral in charge having 1 electron for each atom. An Atom that gains or looses an electron loses that perfect ratio. It is positive is electrons are loss and negative if electrons are gained.

3 0
3 years ago
HCI + NaOH --> NaCl + H 20
Vikki [24]

Answer:

option D is correct

Explanation:

no of moles in 3 grams of HCL=3/36=0.08

if 1 mole of HCL require 1 mole of NaOH then 0.08 moles required 0.08 moles of NaOH

mass of 0.08 moles of NaOH=moles*molar mass=0.08*40=3.2 grams

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6 0
2 years ago
Determine the LIMITING reactant in the following balanced equation:
padilas [110]

Answer:

KBr is limiting reactant.

Explanation:

Given data:

Mass of  KBr =4g

Mass of Cl₂ = 6 g

Limiting reactant = ?

Solution:

Chemical equation:

2KBr + Cl₂      →    2KCl + Br₂

Number of moles of KBr:

Number of moles = mass/molar mass

Number of moles = 4 g/ 119 gmol

Number of moles = 0.03 mol

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 6 g/ 70 gmol

Number of moles = 0.09 mol

Now we will compare the moles of reactant with product.

              KBr            :            KCl

                2              :              2

            0.03            :            0.03

             KBr            :              Br₂

                2             :               1

             0.03           :          1/2×0.03= 0.015

               Cl₂             :            KCl

                 1              :              2

            0.09            :           2/1×0.09 = 0.18

               Cl₂             :              Br₂

                1              :               1

             0.09           :            0.09

Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂  is present in excess.

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