2C_6H_14 + 19O_2 → 12CO_2 + 14H_2O
<em>Step 1</em>. Write the <em>condensed structural formula</em> for 2,3-dimethylbutane.
(CH_3)_2CHCH(CH_3)_2
<em>Step 2</em>. Write the <em>molecular formula</em>.
C_6H_14
<em>Step 3</em>. Write the <em>unbalanced chemical equation</em>.
C_6H_14 + O_2 → CO_2 + H_2O
<em>Step 4</em>. Pick the <em>most complicated-looking formula</em> (C_6H_14) and balance its atoms (C and H).
<em>1</em>C_6H_14 + O_2 → <em>6</em>CO_2 + <em>7</em>H_2O
<em>Step 5</em>. Balance the <em>remaining atoms</em> (O).
1C_6H_14 + (<em>19/2</em>)O_2 → 6CO_2 + 7H_2O
Oops! <em>Fractional coefficients</em>!
<em>Step 6</em>. <em>Multiply all coefficients by a number</em> (2) to give integer coeficients..
2C_6H_14 + 19O_2 → 12CO_2 + 14H_2O
1) Zn + 2 HCl = ZnCl2 + H2 ( <span>single replacement )
2) </span>2 NaCl + F2 = 2 NaF + Cl2 ( <span>single replacement )
3) </span>2 AlBr3 + 3 K2SO4 = 6 KBr + Al2(SO4)3 ( <span>double replacement )
4) </span>2 K + MgBr2 = 2 KBr + Mg ( <span>single replacement )
Answer 3
hope this helps!</span>
Answer:
The area of the given rectangular index card = <u>9677.4 mm²</u>
Explanation:
Area is defined as the space occupied by a two dimensional shape or object. The SI unit of area is square metre (m²).
<u>The area of a rectangle</u> (A) = length (l) × width (w)
Given dimensions of the rectangle: Length (l) = 5.0 inch, Width (w) = 3.0 inch
Since, 1 inch = 25.4 millimetres (mm)
Therefore, l = 5 × 25.4 = 127 mm, and w = 3 × 25.4 = 76.2 mm
Therefore, <u>the area of the given rectangular index card</u> = A= l × w = 127 mm × 76.2 mm = <u>9677.4 mm²</u>
Answer:
λ = 6.5604 x 1016 nm
Explanation:
Given Data:
The energy of the red line in Hydrogen Spectra = 3.03 x 10-19
Formula to calculate Wave length
E= hv
Where E is Energy
h is Planks Constant = 6.626 x 10–34 J s
v is frequency
In turn
v= c/ λ
where c is speed of light = 3.00 x 108 m s–1
λ is wavelength = to find
Solution:
Formula to be Used:
E= hv………………………… (1)
Putting the value v in equation 1
E= h c/ λ…………………… (2)
Put the value in equation 2
3.03 x 10-19 J = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) / λ ……………………….(3)
By rearranging equation 3
λ = (6.626 x 10–34 J s) x (3.00 x 108 m s–1) /3.03 x 10-19 J
λ = 6.5604 x 107 m
The answer is in “m”
So we have to convert it into nm
So for this to convert “m” to “nm” multiply the answer with 109
λ = 6.5604 x 107 x 109
λ = 6.5604 x 1016 nm
Separa los metales de los no metales. Agregame como amiga, saludos.
