There are TWO atoms in one molecule of hydrogen.
Answer:
62.98 % of the sample of hydrate is water
Explanation:
Step 1: Data given
Mass of the sample of a hydrate of sodium carbonate (Na2CO3) = 2.026 grams
After heating, the mass of the sample is 0.750 g
Molar mass H2O = 18.02 g/mol
Step 2: Calculate mass of water
Mass water = mass of hydrate - mass of sample after heating
Mass water = 2.026 grams - 0.750 grams
Mass water = 1.276 grams
Step 3: Calculate mass % percent of water
Mass % of water = (mass of water / total mass hydrate) * 100 %
Mass % of water = (1.276 grams / 2.026 grams) *100 %
Mass % of water = 62.98 %
62.98 % of the sample of hydrate is water
Answer:
3AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄
The coefficients are 3, 1, 3, 1
Explanation:
From the question given above, the following data were:
Silver chloride reacts with sodium phosphate to yield sodium chloride and silver phosphate. This can be written as follow:
AgCl + Na₃PO₄ —> NaCl + Ag₃PO₄
The above equation can be balanced as follow:
AgCl + Na₃PO₄ —> NaCl + Ag₃PO₄
There are 3 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 3 in front of NaCl as shown below:
AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄
There are 3 atoms of Cl on the right side and 1 atom on the left. It can be balance by putting 3 in front of AgCl as shown below:
3AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄
Thus, the equation is balanced.
The coefficients are 3, 1, 3, 1
The only one that I can do without google is 47. Sorry that I can't answer the others. The answer to 47 is this: you know that the western side of the hill has the steepest slope because the ovals showing altitude are way closer together. The closer the circles/ovals are, the steeper the slope is.
Sorry if this doesn't help much, but I answered what I could without cheating.
Foxeslair
Answer:
Answers are in the explanation
Explanation:
Equlibrium of HF in H₂O is:
HF + H₂O ⇄ F⁻ + H₃O⁺
Now, the KOH reacts with HF, thus:
KOH + HF → F⁻ + H₂O
<em>That means after reaction, concentration of HF decrease increasing F⁻ concentration.</em>
Now, seeing the equilibrium, as moles of HF decrease and F⁻ moles increase, the equilibrium will shift to the left decreasing H₃O⁺ concentration.
For the statements:
A. The number of moles of HF will increase. <em>FALSE</em>. HF react with KOH, thus, moles of HF decrease
B. The number of moles of F- will decrease. <em>FALSE</em>. The reaction produce F⁻ increasing its moles.
C. The equilibrium concentration of H₃O⁺ will increase. <em>FALSE. </em>The equilibrium shift to the left decreasing concentration of H₃O⁺
D. The pH will decrease. <em>FALSE</em>. As the H₃O⁺ concentration decrease, pH will increase
E. The ratio of [HF] / [F-] will remain the same. <em>FALSE</em>. Because moles of HF are decreasing whereas F- moles are increasing changing, thus, ratio.
