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Tju [1.3M]
3 years ago
9

Identify the element that has 23 protons

Chemistry
2 answers:
barxatty [35]3 years ago
3 0

Answer:

Vanadium

Explanation:

Brums [2.3K]3 years ago
3 0

Answer:

Vanadium

Explanation:

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Select the correct term to complete each sentence.
leva [86]

Answer: 1. halve

2. halve

3. double

Explanation:

The relationship between wavelength and energy of the wave follows the equation:

E=\frac{hc}{\lambda}

E= energy

\lambda = wavelength of the wave

h = Planck's constant  

c = speed of light

Thus as wavelength and energy have inverse realation, when wavelength will halve , energy will double.

2. The between wavenumber and energy of the wave follows the equation:

E=h\times c\times \bar{\nu}

E= energy

\bar{\nu}\nubar = wavenumber of the wave

h = Planck's constant  

c = speed of light

Thus as wavenumber and energy have direct relation, when wavenumber will halve , energy will be halved.

3. The relationship between energy and frequency of the wave follows the equation:

E=h\times \nu

where

E = energy

h = Planck's constant  

\nu = frequency of the wave

Thus as frequency and energy have direct realation, when frequency will double , energy will double.

3 0
3 years ago
The combustion of glucose (c6h12o6) with oxygen gas produces carbon dioxide and water. this process releases 2803 kj per mole of
V125BC [204]

Answer:- 335 kcal of heat energy is produced.

Solution:- The balanced equation for the combustion of glucose in presence of oxygen to give carbon dioxide and water is:

C_6H_1_2O_6+6O_2\rightarrow 6CO_2+6H_2O

From given info, 2803 kJ of heat is released bu the combustion of 1 mol of glucose. We need to calculate the energy produced when 3.00 moles of oxygen react with excess of glucose.

We could solve this using dimensional analysis as:

3.00mol O_2(\frac{1mol glucose}{6mol O_2})(\frac{2803 kJ}{1mol glucose})

= 1401.5 kJ

Now, let's convert kJ to kcal.

We know that, 1kcal = 4.184kJ

So, 1401.5kJ(\frac{1kcal}{4.184kJ})

= 335 kcal

Hence, 335 kcal of heat energy is produced by the use of 3.00 moles of oxygen gas.


8 0
3 years ago
A blacksmith making a tool heats 525 grams of steel to 1230°C. After hammering the steel, she places it into a bucket of water t
Simora [160]
C
WELL HOPE THIS CAN HELP U
6 0
3 years ago
Erbium (Er) is element 68 on the periodic table. A sample contains 5.04×1023 atoms of Er. Calculate the amount of Er.
Korvikt [17]

Answer:

139.98 g to nearest hundredth.

Explanation:

Using Avogadro's Number:

One mole (167.26 g) of Erbium  equates to  6.022141 * 10^23 atoms.

So 5.04 * 10^23 = 167.26 * 5.04/6.022141

= 139,98 g.

3 0
3 years ago
HELP!!!!!!!!!!!!!!!!!!! 100 POINTSSSSSSSSSSSSSS
e-lub [12.9K]

Answer:

<em><u>The three-dimensional region of space that indicates where there is a high probability of finding an electron.</u></em>

7 0
3 years ago
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