Question: Baking a Cake Without Flour.
Hypothesis: I think that when I remove the flour from the standard cake recipe, I'll end up with a flat but tasty cake.
Procedure: I baked two cakes during my experiment. For my control, I baked a cake following a normal recipe. I used the Double Fudge Cake recipe on page 292 of the Betty Crocker Cookbook. For my experimental cake, I followed the same recipe but left out the flour. I first obtained a 2-quart mixing bowl.
Results: My control cake, which I cooked for 25 minutes, measured 4 cm high. Eight out of ten tasters that I picked at random from the class found it to be an acceptable dessert. After 25 minutes of baking, my experimental cake was 1.5 cm high and all ten tasters refused to eat it because it was burnt to a crisp.
What did I learn?/Conclusion: Since the experimental cake burned, my results did not support my hypothesis. I think that the cake burned because it had less mass, but cooked for the same amount of time. I propose that the baking time be shortened in subsequent trials.
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I hope this helped :))
It is c: a conductor that operates only at low temperatures
Answer:
draw the carbon chain is containing 6 carbon then attach the aldehyde group with edge carbon in chain then put the put double bond at 3 no. carbon
Answer:
11.31g NaClO₂
Explanation:
<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>
It is possible to answer this question using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>
You can change the concentration of the substance if you write the moles of the substances:
[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>
Replacing in H-H expression, as the pH you want is 1.45:
1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]
0.1251 = Moles NaClO₂
As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:
0.1251 moles NaClO₂ ₓ (90.44g / mol) =
<h3>11.31g NaClO₂</h3>
Answer:
Yes. Weight is the product of mass times gravitational acceleration. So all you have to do is vary the gravitational field and you vary weight.
Explanation:
