Given mass of tungsten, W = 415 g
Molar mass of tungsten, W = 183.85 g/mol
Calculating moles of tungsten from mass and molar mass:
Answer:
protista, fungi, plant, animal
Explanation:
this is the answer
Answer:
3.49 g
Explanation:
The mass is the product of volume and density:
(8.96 g/cm³)(0.39 cm³) ≈ 3.49 g
The mass of a pure-copper penny would be 3.49 g.
<span>Answer: 0.094%
</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />
<span>Only the ionization of the formic acid is the important part.
</span><span />
<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />
<span>2) Mass balance:
</span><span />
<span> HCOOH(aq) HCOO⁻(aq) H⁺(aq).
Start 0.311 0.189
Reaction - x +x +x
Final 0.311 - x 0.189 + x x
3) Acid constant equation:
</span><span />
<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />
<span>= (0.189 + x )x / (0.311 - x) = 0.000177
4) Solve the equation:
You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />
<span>With that approximation the equation to solve becomes:
</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M
5) With that number, the percent of ionization (alfa) is:
</span><span />
<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
<span></span><span />
Missing question:
I. switching from granulated sugar to pow-dered sugar.
II. switching from powdered sugar to gran-ulated sugar.
III. warming up the sulfuric acid.
IV. putting the sulfuric acid container in an ice bath before use.
1. IV only
2. II only
3. III only
4. I and IV
5. I and III.
6. I only.
7. II and III
<span>8. II and IV.
Answer is: 5. </span>I and III.
If the surface area and temperature are increased, speed of reaction also increased<span>.</span>
