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algol13
1 year ago
5

I NEED HELP ASAP PLEASE

Chemistry
1 answer:
AnnyKZ [126]1 year ago
4 0

One molecule of ammonia is composed of two atoms of nitrogen and three atoms of hydrogen. Option B.

<h3>What is an equation?</h3>

The term chemical equation has to do with the presentation of a chemical reaction on paper in a way that it can be easily understood. It is easy to write an equation to show what is going on in a reaction system.

Now we have the reactions as shown in the question. In this reaction which is the synthesis of ammonia and occurs industrially in the Haber process. The statement that is not true is that; one molecule of ammonia is composed of two atoms of nitrogen and three atoms of hydrogen. Option B.

Learn more about chemical equation:brainly.com/question/28294176

#SPJ1

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6) (a) Calculate the absorbance of the solution if its concentration is 0.0278 M and its molar extinction coefficient is 35.9 L/
Anvisha [2.4K]

Answer:

6) (a) 0.499; (b) 31.7 %

7) 0.15

Explanation:

6) (a) Absorbance

Beer's Law is

A = \epsilon cl\\A = \text{35.9 L&\cdot$mol$^{-1}$cm$^{-1}$} $\times$ 0.0278 mol$\cdot$L$^{-1} \times $ 0.5 cm = \mathbf{0.499}

(b) Percent transmission

A = \log {\left (\dfrac{1}{T}}\right)}\\\\\%T = 100T\\\\T = \dfrac{\%T}{100}\\\\\dfrac{1}{T} = \dfrac{100 }{\%T}\\\\A = \log \left(\dfrac{100 }{\%T} \right ) = 2 - \log \%T\\\\0.499 = 2 - \log \%T\\\\\log \%T = 2 - 0.499 = 1.501\\\\\%T = 10^{1.501} = \mathbf{31.7}

7) Absorbance

A = \log \left (\dfrac{I_{0}}{I} \right ) = \log \left (\dfrac{I_{0}}{0.70I_{0}} \right ) = \log \left (\dfrac{1}{0.70} \right ) = -\log(0.70) = \mathbf{0.15}}

8 0
3 years ago
What is the oxidation number of carbon in the compound carbon dioxide, CO2?
Anna35 [415]

Answer:

the oxidation number is 4

7 0
2 years ago
A hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:
kupik [55]

hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.hydrogen-like ion is an ion containing only one electron. The energy of the electron in a hydrogen-like ion is given by:

En = −(2.18 × 10^−18J) Z^2 ( 1/n^2 )

where n is the principal quantum number and Z is the atomic number of the element. Plasma is a state of matter consisting of positive gaseous ions and electrons. In the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as Hg80+. Use the equation above to calculate the energy required for the last ionization step.

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6 0
2 years ago
A gray element that borders on the zigzag line of the periodic table, is ductile and malleable, but is not a very good conductor
frez [133]
The answer is non metal D
6 0
2 years ago
A 7.06% aqueous solution of sodium bicarbonate has a density of 1.19g/mL at 25°C what is the molarity and molality of the soluti
lilavasa [31]

<em>c</em> = 1.14 mol/L; <em>b</em> = 1.03 mol/kg

<em>Molar concentration </em>

Assume you have 1 L solution.

Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)

= 1190 g solution

Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)

= 84.01 g NaHCO3

Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)

= 1.14 mol NaHCO3

<em>c</em> = 1.14 mol/1 L = 1.14 mol/L

<em>Molal concentration</em>

Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg

<em>b</em> = 1.14 mol/1.106 kg = 1.03 mol/kg

5 0
3 years ago
Read 2 more answers
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