All categories

2 years ago

Tables that organize mortality data for a year are called

Chemistry

2 answers:

Sloan [31]2 years ago

7 0

Answer:

A cohort, or generation, life table is based on, or represents, mortality experience over the entire lifetime of a cohort of persons born during a relatively short period of time, usually one year.

Explanation:

Anna007 [38]2 years ago

7 0

• Tables that organize mortality data for a year are called : A Life Table

You might be interested in

How and why do ionic bonds form? Check all of the boxes that apply. Ionic bonds form between metal atoms and other metal atoms.

sammy [17]

2 Ionic bonds form between metal atoms and nonmetal atoms.

4 The less electronegative atoms transfers one or more electrons to the more electronegative atom.

5 The metal atom forms a cation and the nonmetal atom forms an anion.

7 The attraction between ions with an opposite charge forms an ionic bond.

8 0

2 years ago

Read 2 more answers

When 5.00 g of FeCl3 . xH2O are heated, 2.00 g of H20 are driven off. Find the chemical formula and the name of the hydrate

cricket20 [7]

Answer:

x551x6x255

Explanation:

if its wrong sorry

3 0

2 years ago

Read 2 more answers

One mole of an ideal gas in a closed system, initially at 25°C and 10 bar, is first expanded adiabatically, then heated isochori

Igoryamba

Answer:

$P_2=0.398bar=39800Pa$

$T_2=118.7K\\$

$Q=-3729.9J$

$W=-61753.24J$

Δ $U_T=0J$

Δ $H_T=0J$

Explanation:

Hello,

At the first state, the molar volume is:

$v_1=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^6Pa}=2.48x10^{-3}m^3$

The volume in both the second and third state:

$v_2=v_3=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^5Pa}=2.48x10^{-2}m^3$

Now, as it is about an adiabatic process, one remembers the following relationships:

$PV^\alpha =K\\TV^{\alpha-1}\\\alpha=\frac{Cp}{Cv}=\frac{7/2R}{5/2R}=1.4$

- Next, for the aforesaid volumes and the first pressure, one computes the second pressure as:

$P_2=\frac{P_1V_1^\alpha }{V_2^\alpha} =\frac{10bar*(2.48x10^{-3}m^3)^{1.4}}{(2.48x10^{-2}m^3)^{1.4}} =0.398bar=39800Pa$

- And the temperature:

$T_2=\frac{T_1V_1^{\alpha-1}}{V_2^{\alpha-1}} =\frac{298.15K*(2.48x10^{-3}m^3)^{1.4-1}}{(2.48x10^{-2}m^3)^{1.4-1}} =118.7K\\$

- Q:

It is clear that the heat for the first process is 0 as it is adiabatic, but for the second one, it is computed as:

$Q_2=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J$

Then the total heat:

$Q=Q_1+Q_2=0-3729.9J=-3729.9J$

- The work for the first process is:

$W_1=\frac{P_2V_2-P_1V_1}{1-\alpha }=\frac{39800Pa*2.48x10^{-3}m^3-1x10^6Pa*2.48x10^{-2}m^3}{0.4} \\W_1=-61753.24J$

It is clear that the second process is isochoric, so the work here is zero, thus, the total work is:

$W=W_1+W_2=-61753.24J+0J=-61753.24J$

- For the two processes, ΔU becomes the same value since the system returns to the initial temperature, so ΔU total is 0, thus, for each process, one's got:

$U_1=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J\\U_2=nCv(T_3-T_2)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=3729.9J\\$

- Finally, the total enthapy is also 0 due to same aforesaid reason, thus, each enthalpy is:

$H_1=nCp(T_2-T_1)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-5221.86J\\H_2=nCv(T_3-T_2)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=5221.86J\\$

Best regards.

8 0

2 years ago

A large flask is evacuated and weighed, filled with argon gas, and then reweighed. When reweighed, the flask is found to have ga

Elden [556K]

Answer:

100.52

Explanation:

from the ideal gas equation PV=nRT

for a given container filled with any ideal gas P and V remains constant.So T is also constant.R is as such a constant.

So n i.e no of moles will also be constant.

no of moles of Ar=3.224/40=0.0806

no of moles of unknown gas=0.0806

molecular wt of unknown gas=8.102/0.0806=100.52

8 0

2 years ago

The Kp for the reaction below is 1.49 × 108 at 100.0°C:

zhenek [66]

In an equilibrium mixture of the three gases, PCO = PCl2 = 2.22 × 10-4 atm. The partial pressure of the product, phosgene (COCl2), is kp=(COCl2)/(CO)(Cl2) which is $1.48*10^=x/(2.24*10-4)^2$ . So, the correct answer is <span>A) 7.34.</span>

8 0

2 years ago

Read 2 more answers