<span>0.0165 m
The balanced equation for the reaction is
AgNO3 + MgCl2 ==> AgCl + Mg(NO3)2
So it's obvious that for each Mg ion, you'll get 1 AgCl molecule as a product. Now calculate the molar mass of AgCl, starting with looking up the atomic weights.
Atomic weight silver = 107.8682
Atomic weight chlorine = 35.453
Molar mass AgCl = 107.8682 + 35.453 = 143.3212 g/mol
Now how many moles were produced?
0.1183 g / 143.3212 g/mol = 0.000825419 mol
So we had 0.000825419 moles of MgCl2 in the sample of 50.0 ml. Since concentration is defined as moles per liter, do the division.
0.000825419 / 0.0500 = 0.016508374 mol/L = 0.016508374 m
Rounding to 3 significant figures gives 0.0165 m</span>
<span>A pulse with an amplitude of 3+ would be considered as increased.
Peripheral Pulse Assessment Grading System is measured in 0 - 3 Scale.
0 = absent
1+ = Weak/thready pulse
2+ Normal Pulse
3+ = Full, firm pulse.
from the above scale we can find that the 3+ reading shows that the pulse is increased.</span>
Answer:
71.7 L
Explanation:
Using the ideal gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/Kmol)
T = temperature (K)
According to the information provided in this question;
P = 1 atm (STP)
V = ?
n = 3.2mol
T = 273K (STP)
Using PV = nRT
V = nRT/P
V = 3.2 × 0.0821 × 273/1
V = 71.7 L
Answer:
53.6 grams of silver chloride was produced.
Explanation:
Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.
This also means that total mass on the reactant side must be equal to the total mass on the product side.
Mass of silver nitrate = 50.0 g
Mass of hydrogen chloride = 50.0 g
Mass of silver chloride = x
Mass of nitric acid = 46.4 g
Mass of silver nitrate + Mass of hydrogen chloride =
Mass of silver chloride + Mass of nitric acid
[te]50.0 g+50.0 g=x+46.4 g[/tex]
53.6 grams of silver chloride was produced.
True.
SF are used for simplifying figures in a measurement to produce a more accurate reading.
