Answer: 55.84L
Explanation: Please see attachment for explanation.
Answer:
Scandium
Titanium
Vanadium
Chromium
Manganese
Iron
Cobalt
Nickel
Copper
Zinc
Yttrium
Zirconium
Niobium
Molybdenum
Technetium
Ruthenium
Rhodium
Palladium
Silver
Cadmium
Lanthanum
Hafnium
Tantalum
Tungsten
Rhenium
Osmium
Iridium
Platinum
Gold
Mercury
Actinium
Rutherfordium
Dubnium
Seaborgium
Bohrium
Hassium
Meitnerium
Darmstadtium
Roentgenium
Copernicium
Explanation:
all of those are transition metals lol
The pH of a 0.0115 m aqueous formic acid solution is mathematically given as
pH=2.8424
This is further explained below.
<h3>What is the ph of a 0.0115 m aqueous formic acid solution?</h3>
Generally, the equation for the chemical equation is mathematically given as
HCOOH H^+ + HCOO
![&\left[\mathrm{H}^{+}\right]=\mathrm{C \alpha}\\\\&=0.125 \times 0.0115\\\\&=1.4375 \times 10^{-9}\\\\&P=-\log \left[H^{+}\right]\\\\&=-\log \left[1.4375 \times 10^{-3}\right]\\\\&P H=2.8424](https://tex.z-dn.net/?f=%26%5Cleft%5B%5Cmathrm%7BH%7D%5E%7B%2B%7D%5Cright%5D%3D%5Cmathrm%7BC%20%5Calpha%7D%5C%5C%5C%5C%26%3D0.125%20%5Ctimes%200.0115%5C%5C%5C%5C%26%3D1.4375%20%5Ctimes%2010%5E%7B-9%7D%5C%5C%5C%5C%26P%3D-%5Clog%20%5Cleft%5BH%5E%7B%2B%7D%5Cright%5D%5C%5C%5C%5C%26%3D-%5Clog%20%5Cleft%5B1.4375%20%5Ctimes%2010%5E%7B-3%7D%5Cright%5D%5C%5C%5C%5C%26P%20H%3D2.8424)
Read more about chemical equation
brainly.com/question/28294176
#SPJ1
Answer:
A veinlike deposit, usually metalliferous.
Any body of ore set off from adjacent rock formations.
A rich supply or source
Answer:
4.6 × 10²³ molecules:
Step-by-step solution
You will need a balanced equation with masses, moles, and molar masses, so let's gather the information in one place:
M_r: 22.99
2Na + 2H₂O ⟶ 2NaOH + H₂
m/g: 35
1. Calculate the <em>moles of Na
</em>
Moles of Na = 35 g Na × (1 mol Na/22.99 g Na)
Moles of Na = 1.52 mol Na
2. Calculate the <em>moles of H₂
</em>
Moles of H₂ = 1.52 mol Na × (1 mol H₂/2 mol Na)
Moles of H₂= 0.761 mol H₂
3. Calculate the molecules of H₂
6.022 × 10²³ molecules H₂ = 1 mol H₂
Molecules of H₂ = 0.761 × (6.022 × 10²³
/1)
Molecules of H₂ = 4.6 × 10²³ molecules H₂
The reaction forms 4.6 × 10²³ molecules of H₂.
