Answer:
Wind
Explanation:
It's impossible for them to travel in a wave.
Answer:
THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO
Explanation:
The steps involved in calculating the empirical formula of this substance in shown in the table below:
Element Carbon Hydrogen Nitrogen Oxygen
1. % Composition 40.66 8.53 23.72 27.09
2. Mole ratio =
%mass/ atomic mass 40.66/12 8.53/1 23.72/14 27.09/16
= 3.3883 8.53 1,6943 1.6931
3. Divide by smallest
value (0.6931) 3.3883/1.6931 8.53/1.6931 1.6943/1.6931 1.6931/1.6931
= 2.001 5.038 1.0007 1
4. Whole number ratio 2 5 1 1
The empirical formula = C2H5NO
Answer:
I believe the answer is "b". "During the experiment, the scientist has only one element, or variable, that is changed to test the hypothesis."
Explanation:
I remember from last year but I'm not totally sure. Good luck!
Answer:
21.10g of H2O
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2C7H14 + 21O2 —> 14CO2 + 14H2O
From the balanced equation above, 2L of C7H14 produced 14L of H2O.
Therefore, 3.75L of C7H14 will produce = (3.75 x 14)/2 = 26.25L of H2O.
Next, we shall determine the number of mole of H2O that will occupy 26.25L at stp. This is illustrated below:
1 mole of a gas occupy 22.4L at stp
Therefore, Xmol of H2O will occupy
26.25L i.e
Xmol of H2O = 26.25/22.4
Xmol of H2O = 1.172 mole
Therefore, 1.172 mole of H2O is produced from the reaction.
Next, we shall convert 1.172 mole of H2O to grams. This is illustrated below:
Number of mole H2O = 1.172 mole
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H2O =..?
Mass = mole x molar mass
Mass of H2O = 1.172 x 18
Mass of H2O = 21.10g
Therefore, 21.10g of H2O is produced from the reaction.
Answer:
V KOH = 41 mL
Explanation:
for neutralization:
- ( V×<em>C </em>)acid = ( V×<em>C </em>)base
∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L
∴ V H2SO4 = 41 mL = 0.041 L
∴ <em>C</em> KOH = 0.0050 N = 0.0050 eq-g/L
∴ E KOH = 1 eq-g/mol
⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L
⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH
⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)
⇒ V KOH = 0.041 L
