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3 years ago

How do we use properties of matter in our every day lives

1 answer:

6 0

For example, water. In its liquid state we drink it , in its solid state (ice ) we use it to make our drinks colder or t numbs pain when wrapped in cloth and it’s gas state ( water vapour) we steam vegetables. In conclusion many different properties of matter are used for various purposes

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An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its

german

Hello!

First you need to calculate q
delta U is change in internal energy 

delta U = q + w 
q is heat and w work done 
here work was done by the system means energy leaving the system so w is negative 

delta U = q + w 

q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ 

q = m x c x delta T 

7211 J = 80.0 g x c x (225-25) °C 

c = 0.451 J /g °C

Hope this Helps! Have A Wonderful Day! :)

5 0

3 years ago

electric current is passed through water, it decomposes inti hydrogen and oxygen. write full balanced chemical equation for the

Tems11 [23]

Answer:

2H2O ------------------] 2H2 + O2

6 0

3 years ago

The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A

s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $m^3 /d$

$= 80 \times 10^5 \ I/d$

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L$

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d$

If 60 mg of alum $[Al_2(SO_4)_3.14 H_2O]$ required for one litre of the water treatment.

So Alum required for $80 \times 10^5 \ I/d$

$= 80 \times 15^5 \ I/d \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d$

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $= 60 \times 0.234 \text{ of alum ppt. per litre}$

$= 14.04$ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

= 112.32 kg/d ppt of alum

Daily total solid load is $= 144 \ kg/d + 112.32 \ kg/d$

= 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234$

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-}$

After addition, the aluminium hydroxide pH of water will increase due to increase in $OH^-$ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm$

And the pH is reduced into the range of 5.9 to 6.4

6 0

3 years ago

How many valence electrons are in each atom? a. potassium b. carbon c. magnesium d. oxygen

Alona [7]

A. 1 valence electron
b. 4 valence electrons
c. 2 valence electrons
d. 6 valence electrons

8 0

2 years ago

What is the half-life of A? What will the pressure, initially 32.1 kPa, at

kodGreya [7K]

Answer:

a) 32.09 kPa

b) 32.09 kPa

Explanation:

Given data:

rate constant $= 3.56\times 10^{-7} s^{-1}$

initial pressure is = 32.1 kPa

half life of A is calculated as

$t_{1/2} = \frac{ln 2}{k}$

$t_{1/2} = \frac{ln 2}{3.56\time 10^{-7}}$

$t_{1/2} = = 1.956 \times 10^6 s$

for calculating pressure we have follwing expression

$ln p = ln P_o - kt$

$P =P_o e^{-kt}$

a) $P = 32.1 e^{-3.56\times 10^{-7} \times 10} = 32.09 kPa$

b) $P = 32.1 e^{-3.56\times 10^{-7} \times 10\times 60} = 32.09 kPa$

5 0

3 years ago