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
Answer:
The enthalpy is increased by the increased heat of the reaction.
Explanation:
In this reaction, as the transition from solid to liquid state, enthalpy increases, that is, the heat applied to change its state is greater and this increases, reaching a mayor disorder.
If the reaction increases its heat, and a liquid state passes, the enthalpy increases, therefore the disorder also and the entropy will also be increased.
The reactants, products, coefficients, subscripts. ( I forgot the rest lol)
moles NaOH = c · V = 0.1973 mmol/mL · 29.43 mL = 5.806539 mmol
moles H2SO4 = 5.806539 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 2.9032695 mmol
Hence
[H2SO4]= n/V = 2.9032695 mmol / 32.42 mL = 0.08955 M
The answer to this question is [H2SO4] = 0.08955 M
Answer:
b. potassium.
Explanation:
Potassium-sparing diuretics and salt substitutes are diuretics that eliminate salt and water but save potassium. They act by inhibiting the conducting sodium channels in the collecting tubule, such as amiloride and triamterene, or by blocking aldosterone, such as spironolactone.
Concomitant use of potassium-sparing diuretics together with salt substitutes may result in dangerously high blood levels of serum potassium. For this reason, it is important to consult a physician before taking these substances at the same time to avoid potential problems with potassium accumulation.
