Majority can such as hydrogen H and oxygen O forming water H2O but uranium having extra neutrons to form plutonium and beyond simply can't because it will not last for a fraction of a second or spiral out of control and violently react like the little boy and the fat man (a uranium and plutonium nuclear weapon) so yes and no
True....
Explanation
Cells use oxygen to release energy during photosynthesis/chlorophyll/cellular respiration. For photosynthesis, the starting materials are water and carbon dioxide, and the products are sugar and oxygen. For cellular respiration, the starting materials are oxygen and sugar, and the products are carbon dioxide and water.
Answer:
THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K
Explanation:
Mass of alloy = 33 g
Initial temperature of alloy = 93°C
Mass of water = 50 g
Initail temp. of water = 22 °C
Heat capacity of calorimeter = 9.20 J/K
Final temp. = 31.10 °C
specific heat of alloy = unknown
specific heat capacity of water = 4.2 J/g K
Heat = mass * specific heat * change in temperature = m c ΔT
Heat = heat capcity * chage in temperature = Δ H * ΔT
In calorimetry;
Heat lost by the alloy = Heat gained by water + Heat of the calorimeter
mc ΔT = mcΔT + Heat capacity * ΔT
33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)
33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1
2042.7 C = 1911 + 83,72
C = 1911 + 83.72 / 2042.7
C = 1994.72 /2042.7
C =0.9765 J/g K
The specific heat of the alloy is 0.9765 J/ g K
Explanation:
To answer this question, we'll need to use the Ideal Gas Law:
p
V
=
n
R
T
,
where
p
is pressure,
V
is volume,
n
is the number of moles
R
is the gas constant, and
T
is temperature in Kelvin.
The question already gives us the values for
p
and
T
, because helium is at STP. This means that temperature is
273.15 K
and pressure is
1 atm
.
We also already know the gas constant. In our case, we'll use the value of
0.08206 L atm/K mol
since these units fit the units of our given values the best.
We can find the value for
n
by dividing the mass of helium gas by its molar mass:
n
=
number of moles
=
mass of sample
molar mass
=
6.00 g
4.00 g/mol
=
1.50 mol
Now, we can just plug all of these values in and solve for
V
:
p
V
=
n
R
T
V
=
n
R
T
p
=
1.50 mol
×
0.08206 L atm/K mol
×
273.15 K
1 atm
= 33.6 L
this is not the answer but it will help you
do by the formula it is on the answer
