94.6 g. You must use 94.6 g of 92.5 % H_2SO_4 to make 250 g of 35.0 % H_2SO_4.
We can use a version of the <em>dilution formula</em>
<em>m</em>_1<em>C</em>_1 = <em>m</em>_2<em>C</em>_2
where
<em>m</em> represents the mass and
<em>C</em> represents the percent concentrations
We can rearrange the formula to get
<em>m</em>_2= <em>m</em>_1 × (<em>C</em>_1/<em>C</em>_2)
<em>m</em>_1 = 250 g; <em>C</em>_1 = 35.0 %
<em>m</em>_2 = ?; _____<em>C</em>_2 = 92.5 %
∴ <em>m</em>_2 = 250 g × (35.0 %/92.5 %) = 94.6 g
Answer:
B
Explanation:
it shows that the solution is neutral.
I hope this helps
Answer:
3 P atoms
Explanation:
Al₂P₃ => contains 2 Aluminum ions (2Al⁺³) and 3 Phosphide ions (3P⁻³) ... The ions (charged particles) are from atoms that have lost or gained electrons during the bonding process. So, Al₂P₃ => P⁻³ ions from 3 P atoms.
<span>31.9 grams butane needed to produce 96.7 grams CO2
</span>
Answer:
The three major types of bond are ionic, polar covalent, and covalent bonds. Ionic occurs majorly between metals and non-metals, which allows sharing of electrons to form an ionic compound. Whereas covalent bonding calls for complete transfer of electrons between atoms. Polar covalent bonds have unequaly shared electron-pair between two atoms.
Explanation:
a. Cu (Copper)-<em> ionic bonding
</em>
b. KCl (Potassium Chloride)
- <em>ionic bonding
</em>
c. Si (Silicon)
- <em>covalent bonding
</em>
d. CdTe (Cadmium Telluride)
- <em>polar covalent bonding
</em>
e. ZnTe (Zinc Telluride)- <em>polar covalent bonding
</em>