In an oxoacid such as H2SO4, ionizable hydrogen atoms are those bonded to oxygen

A lab scientist cools a liquid sample of water (2.6 kg) at 0.00°C to -192°C. The water turns to ice as this temperature change o

soldier1979 [14.2K]

Answer: The heat released for the given process is -1892 kJ

Explanation:

The processes involved in the given problem are:

$1.)H_2O(l)(0^oC,273K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(s)(-192^oC,81K)$

Pressure is taken as constant.

To calculate the amount of heat released at same temperature, we use the equation:

$q=m\times L_{f,v}$ ......(1)

where,

q = amount of heat released = ?

m = mass of water/ice

$L_{f,v}$ = latent heat of fusion or vaporization

To calculate the amount of heat released at different temperature, we use the equation:

$q=m\times C_{p,m}\times (T_{2}-T_{1})$ .......(1)

where,

q = amount of heat released = ?

$C_{p,m}$ = specific heat capacity of medium

m = mass of water/ice

$T_2$ = final temperature

$T_1$ = initial temperature

Calculating the heat absorbed for each process:

For process 1:

Converting the latent heat of fusion in J/kg, we use the conversion factor:

1 kJ = 1000 J

So, $(\frac{-334kJ}{1kg})\times (\frac{1000J}{1kJ})=-334\times 10^3J/kg$

We are given:

$m=2.6kg\\L_f=-334\times 10^3J/kg$

Putting values in equation 1, we get:

$q_1=2.6kg\times (-334\times 10^3J/kg)=-868400J$

For process 2:

We are given:

$m=2.6kg\\C_{p,s}=2050J/kg.K\\T_1=273K\T_2=81K$

Putting values in equation 2, we get:

$q_2=2.6kg\times 2050J/kg.K\times (81-(273))^oC\\\\q_2=1023360J$

Total heat absorbed = $q_1+q_2$

Total heat absorbed = $[-868400+(-1023360)]J=-1891760J=-1891.76kJ\approx -1892kJ$

Hence, the heat released for the given process is -1892 kJ

3 0

3 years ago

Which two functional groups react to form an easter linkage?

7nadin3 [17]

Answer: You know that monomers that are joined by condensation polymerization have two functional groups. You also know (from Part 6) that a carboxylic acid and an amine can form an amide linkage, jand a carboxylic acid and an alcohol can form an ester linkage.

5 0

3 years ago

There are two naturally occurring isotopes of bromine. 79Br has a mass of 78.9183 amu. 81Br has a mass of 80.9163 amu. Determine

Ede4ka [16]

Atomic mass of Br = 79.904
Now lets say y% is abundance of 79Br. 
Then abundance of 81Br = (100 - y) 
mass due to 79Br = 78.9183 * y/100 = 0.789183y
mass due to 81Br = 80.9163 x (100 - y)/100 = 0.809163(100 - y) 
Therfore
0.789183y+ 0.809163(100 - y) = 79.904 
0.789183y + 80.9163 - 0.809163y = 79.904 
 - 0.01998y= 79.904 - 80.9163
= - 1.0123 
y = 1.0123/0.01998 = 50.67% 
 79Br = 50.67% 
now
81Br = 100 - 50.67 = 49.33%
hope this helps

4 0

4 years ago

Please help I really need but this time I'm fr I don't have much time ill mark brainliest

FinnZ [79.3K]

Answer:

i fi am not wrong the correct answer is D

Explanation:

5 0

3 years ago

Read 2 more answers

Please help me get the answer to bothe a and b

neonofarm [45]

Answer:

If a metal and metal solution react, the more reactive metal will displace the less reactive metal from solution. If the metal in solution you start with is formed from a more reactive metal than the metal to be added, no reaction will occur.

3 0

2 years ago