Answer:
7.462
Explanation:
Well, every time that the tempurature is increased, the atmspheric pressure is increased by 0.574%. This would then mean that you would have 0.574 times
13. That would then equal 7.462. I hope this helps.
I don't see the options for an answer, so here is a list of all of the transition metals lol
- <em>Scandium</em>
- <em>Titanium</em>
- <em>Vanadium</em>
- <em>Chromium</em>
- <em>Manganese</em>
- <em>Iron</em>
- <em>Cobalt</em>
- <em>Nickel</em>
- <em>Copper</em>
- <em>Zinc</em>
- <em>Yttrium</em>
- <em>Zirconium</em>
- <em>Niobium</em>
- <em>Molybdenum</em>
- <em>Technetium</em>
- <em>Ruthenium</em>
- <em>Rhodium</em>
- <em>Palladium</em>
- <em>Silver</em>
- <em>Cadmium</em>
- <em>Lanthanum</em>
- <em>Hafnium</em>
- <em>Tantalum</em>
- <em>Tungsten</em>
- <em>Rhenium</em>
- <em>Osmium</em>
- <em>Iridium</em>
- <em>Platinum</em>
- <em>Gold</em>
- <em>Mercury</em>
- <em>Actinium</em>
- <em>Rutherfordium</em>
- <em>Dubnium</em>
- <em>Seaborgium</em>
- <em>Bohrium</em>
- <em>Hassium</em>
- <em>Meitnerium</em>
- <em>Darmstadtium</em>
- <em>Roentgenium</em>
- <em>Copernicium p</em>
Answer:
there are two significant figures is the number 8400
Explanation:
Answer:
Mass of solid = 189.141 gram
Explanation:
Given:
Total volume = 93 ml
Mass of liquid = 33.7 gram
Density of liquid = 0.865 g/ml
Density of solid = 3.50 g/ml
Find:
Mass of solid = ?
Computation:
Volume of liquid = Mass of liquid / Density of liquid
Volume of liquid = 33.7 / 0.865
Volume of liquid = 38.9595 ml
Volume of solid = Total volume - Volume of liquid
Volume of solid = 93 - 38.9595
Volume of solid = 54.0405 ml
Mass of solid = Volume of solid × Density of solid
Mass of solid = 54.0405 ml × 3.50 g/ml
Mass of solid = 189.141 gram
Answer is: <span>No, because she did not stop adding base once the color changed.
</span>The endpoint<span> is the point at which the indicator changes colour in a colourimetric </span>titration and that is point when titration must stop or results are going to be wrong, because t<span>he </span><span>equivalence point of titration is not measured right.</span>
