V1/T1 =V2/T2 (using charles law)
V1=6.00
V2=?
T1=273
T2=273
Making V2 the subject of formula the equation then becomes
V2= V1xT2/T1
6.00x263/273=6.0L
The percentage adds up to 100%.
<h3>Percentages</h3>
Mass of mixture = mass of container+mixture - mass of container = 56.779 - 54.558 = 2.221 g
CuCO3 is insoluble in water. Thus:
Mass of CuCO3 = mass of beaker and residue - mass of beaker = 78.875 - 77.575 = 1.300 g
Mass of NaCl = mass of mixture - mass of CuCO3 = 2.221 - 1.300 = 0.921 g
%NaCl (w/w) = weight of NaCl/weight of mixture = 10.921/2.221 = 41.468%
% (w/w) CuCO3 = weight of CuCO3/weight of mixture = 1.300/2.221 = 58.532%
Addition of percentages = 41.468 + 58.532 = 100%
More on percentages can be found here: brainly.com/question/24159063
#SPJ1
Answer:
The density is 0.0187 g/L
Explanation:
First thing to do here is to calculate the Volume of 1 mole of CO2 using the ideal gas equation
Mathematically;
PV = nRT
thus V = nRT/P
what we have are;
n = 1 mole
R is the molar has constant = 0.082 L•atm•mol^-1•K^-1
P is the pressure = 0.0079 atm
T is temperature = 227 K
Substituting these values, we have;
V = nRT/P = (1 * 0.082 * 227)/0.0079
V = 2,356.20 dm^3
This means according to the parameters given in the question, the volume of 1 mole of carbon iv oxide is 2,356.20 dm^3
But this is not what we want to calculate
What we want to calculate is the density
Mathematically, we can calculate the density using the formula below;
density = molar mass/molar volume
Kindly recall that the molar mass of carbon iv oxide is 44 g/mol
Thus the density = 44/2356.20 = 0.018674136321195 which is approximately 0.0187 g/L
To determine the time, we can simply do dimensional analysis from the given values. We are given the distance the fluid travels per sec and we are given the required distance to travel. Therefore, we simply divide the required distance with the rate given. It is important to take note with the units.
t = .01 m / .001 m/s = 10 s
i didn’t mean to put this opps
