Answer:
Commonly available heat-storage materials cannot usually store the energy for a prolonged period. If a solid material could conserve the accumulated thermal energy, then its heat-storage application potential is considerably widened. Here we report a phase transition material that can conserve the latent heat energy in a wide temperature range, T<530 K and release the heat energy on the application of pressure. This material is stripe-type lambda-trititanium pentoxide, λ-Ti3O5, which exhibits a solid–solid phase transition to beta-trititanium pentoxide, β-Ti3O5. The pressure for conversion is extremely small, only 600 bar (60 MPa) at ambient temperature, and the accumulated heat energy is surprisingly large (230 kJ L−1). Conversely, the pressure-produced beta-trititanium pentoxide transforms to lambda-trititanium pentoxide by heat, light or electric current. That is, the present system exhibits pressure-and-heat, pressure-and-light and pressure-and-current reversible phase transitions. The material may be useful for heat storage, as well as in sensor and switching memory device applications.
Explanation:
Answer:
2,375 cans
Explanation:
The strategy here is to use the information given to calculate the lethal dosage contained in the number of cans we will compute.
We know the lethal dosage is
Ld = 10.0 g caffeine
and we also know that the oncentration of caffeine is:
2.85 mg/ oz
So our problem simplifies to calculate how many oz will contain the lethal dose, and then given the ounces per can determine how many cans are required.
First convert the lethal dose in grams to mg:
Ld =( 10 g x 1000 mg ) = 10,000 mg caffeine
10,000 mg x ( 1 Oz / 2.85 mg ) = 28,500 oz
28500 oz x ( 1 can/12 oz ) = 2,375 cans
We could also have calculated it in one step using conversion factors:
Number of cans = 10000 mg x 1 oz/ 2.85 mg x 1 can / oz = 2,375 cans
Answer:
Your strategy here will be to use the molar mass of potassium bromide,
KBr
, as a conversion factor to help you find the mass of three moles of this compound.
So, a compound's molar mass essentially tells you the mass of one mole of said compound. Now, let's assume that you only have a periodic table to work with here.
Potassium bromide is an ionic compound that is made up of potassium cations,
K
+
, and bromide anions,
Br
−
. Essentially, one formula unit of potassium bromide contains a potassium atom and a bromine atom.
Use the periodic table to find the molar masses of these two elements. You will find
For K:
M
M
=
39.0963 g mol
−
1
For Br:
M
M
=
79.904 g mol
−
1
To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements
M
M KBr
=
39.0963 g mol
−
1
+
79.904 g mol
−
1
≈
119 g mol
−
So, if one mole of potassium bromide has a mas of
119 g
m it follows that three moles will have a mass of
3
moles KBr
⋅
molar mass of KBr
119 g
1
mole KBr
=
357 g
You should round this off to one sig fig, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs
mass of 3 moles of KBr
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
360 g
a
a
∣
∣
−−−−−−−−−
Explanation:
<em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em><em> </em><em>3</em><em>6</em><em>0</em><em> </em><em>g</em><em> </em>
