Answer:
Percent error = 1.3%
Explanation:
Given data:
Actual value of boiling point = 100.0 °C
Measured value = 101.3 °C
Percent error = ?
Solution:
Formula:
Percent error = ( Measured value - Actual value / Actual value ) × 100
Now we will put the values in formula.
Percent error = 101.3 °C - 100.0 °C / 100.0 °C × 100
Percent error = 1.3 °C / 100.0 °C × 100
Percent error = 0.013 × 100
Percent error = 1.3%
Thus, percent error is 1.3.
Answer:
First slide:
1) diffusion
2)water
3)high
4)low
5)balance
Second slide:
1)movement
2)without
3)energy
4)high
5)low
6)equilibrium
<u><em>Final slide: </em></u>
<u><em>1)passive</em></u>
<u><em>2)energy </em></u>
<u><em>3)randomly</em></u>
<u><em>4)high</em></u>
<u><em>5)low</em></u>
<u><em>6)equilibruim </em></u>
<u><em>7)permeable </em></u>
<u><em>8)Osmosis</em></u>
<u><em>9)water</em></u>
Hopes this helps! Good luck!
Mass number,atomic number and change i guess
What is [H+] given that the measured cell potential is -0.464 V and the anode reduction ... What half-reaction occurs at the cathode during the electrolysis of molten ... PbO2(s) + 4H+(aq) + SO42-(aq) + 2e- → PbSO4(s) + 2H2O(l); E° = 1.69 V .... For the cell Cu(s)|Cu2+||Ag+|Ag(s), the standard cell potential is 0.46 V. A cell ... hopw this helps
Answer:
87.9%
Explanation:
Balanced Chemical Equation:
HCl + NaOH = NaCl + H2O
We are Given:
Mass of H2O = 9.17 g
Mass of HCl = 21.1 g
Mass of NaOH = 43.6 g
First, calculate the moles of both HCl and NaOH:
Moles of HCl: 21.1 g of HCl x 1 mole of HCl/36.46 g of HCl = 0.579 moles
Moles of NaOH: 43.6 g of NaOH x 1 mole of NaOH/40.00 g of NaOH = 1.09 moles
Here you calculate the mole of H2O from the moles of both HCl and NaOH using the balanced chemical equation:
Moles of H2O from the moles of HCl: 0.579 moles of HCl x 1 mole of H2O/1 mole of HCl = 0.579 moles
Moles of H2O from the moles of NaOH: 1.09 moles of HCl x 1 mole of H2O/1 mole of NaOH = 1.09 moles
From the calculations above, we can see that the limiting reagent is HCl because it produced the lower amount of moles of H2O. Therefore, we use 0.579 moles and NOT 1.09 moles to calculate the mass of H2O:
Mass of H2O: 0.579 moles of H2O x 18.02 g of H2O/1 mole of H2O = 10.43 g
% yield of H2O = actual yield/theoretical yield x 100= 9.17 g/10.43 g x 100 = 87.9%