Answer:
-10778.95 J heat must be removed in order to form the ice at 15 °C.
Explanation:
Given data:
mass of steam = 25 g
Initial temperature = 118 °C
Final temperature = 15 °C
Heat released = ?
Solution:
Formula:
q = m . c . ΔT
we know that specific heat of water is 4.186 J/g.°C
ΔT = final temperature - initial temperature
ΔT = 15 °C - 118 °C
ΔT = -103 °C
now we will put the values in formula
q = m . c . ΔT
q = 25 g × 4.186 J/g.°C × -103 °C
q = -10778.95 J
so, -10778.95 J heat must be removed in order to form the ice at 15 °C.
A wave.
Scientists now recognize that light can behave as both a particle and a wave.
Answer:
A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. The weight/weight % or percent by mass of the solute is :
<u>2.67%</u>
Explanation:
Note : Look at the density of potassium nitrate in water if given in the question.
<u><em>You are calculating </em></u><u><em>weight /Volume</em></u><u><em> not weight/weight % or percent by mass of the solute</em></u>
Here the <u>weight/weight % or percent by mass</u> of the solute is asked : So first convert the<u> VOLUME OF SOLUTION into MASS</u>
Density of potassium nitrate in water KNO3 = 2.11 g/mL
Density = 2.11 g/mL
Volume of solution = 86.4 mL
Mass of Solute = 4.87 g
Mass of Solution = 183.2 g
w/w% of the solute =
w/w%=2.67%
Answer:
Explanation:
El hielo encierra la primavera antes de espolvorear la sal.
Cuando se rocía sal sobre el hielo, disminuye el punto de fusión del hielo 32 ° F a un poco por debajo de 32 °, por lo tanto, se acumula.
A medida que el hielo se vuelve a congelar, encierra la primavera
Answer:
Average atomic mass = 17.5 amu.
Explanation:
Given data:
X-17 isotope = atomic mass17.2 amu, abundance:78.99%
X-18isotope = atomic mass 18.1 amu, abundance 10.00%
X-19isotope = atomic mass:19.1 amu, abundance: 11.01%
Average atomic mass of X = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (78.99×17.2)+(10.00×18.1) +(11.01+ 19.1) /100
Average atomic mass = 1358.628 + 181 +210.291 / 100
Average atomic mass = 1749.919 / 100
Average atomic mass = 17.5 amu.