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3 years ago

Calculate the heat change involved when 2.00 L of water is heated from 20.0/C to 99.7/C in

Chemistry

1 answer:

Pani-rosa [81]3 years ago

5 0

Answer:

666,480 Joules or 669.48 kJ

Explanation:

We are given;

Volume of water as 2.0L or 2000 ml

but, density of water is 1 g/ml

Therefore, mass of water is 2000 g

Initial temperature as 20 °C
Final temperature as 99.7° C

Required to determine the heat change

We know that ;

Heat change = Mass × Temperature change × specific heat

In this case;

Specific heat of water is 4.2 J/g°C

Temperature change is 79.7 °C

Therefore;

Heat change = 2000 g × 79.7 °C × 4.2 J/g°C

= 669,480 Joules 0r 669.48 kJ

Thus, the heat change involved is 666,480 Joules or 669.48 kJ

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The density of water is 1 gm/ml. An object has a mass of 58 grams. What volume must it have in order to float in water?

Keith_Richards [23]

Answer:

58mL

Explanation:

Given parameters:

Density of water = 1g/mL

Mass of object = 58g

Unknown:

The volume the object must have to be able to float in water = ?

Solution:

To solve this problem, we know that the object must have density value equal to that of water or less than that of water to be able to float.

We then set its density to that of water;

Density = $\frac{mass}{volume}$

Volume = $\frac{mass}{density}$

So;

Volume = $\frac{58}{1}$ = 58mL

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3 years ago

How is Carbon returned to the atmosphere Please help image below

Harrizon [31]

Answer:

Carbon is released back into the atmosphere when organisms die, volcanoes erupt, fires blaze, fossil fuels are burned, and through a variety of other mechanisms.Humans play a major role in the carbon cycle through activities such as the burning of fossil fuels or land development.

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3 years ago

Read 2 more answers

n unknown metal is either aluminum, iron or lead. If 150. g of this metal at 150.0 °C was placed in a calorimeter that contains

Nitella [24]

Answer : The metal used was iron (the specific heat capacity is $0.449J/g^oC$ ).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of unknown metal = ?

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of unknown metal = 150 g

$m_2$ = mass of water = 200 g

$T_f$ = final temperature of water = $34.3^oC$

$T_1$ = initial temperature of unknown metal = $150.0^oC$

$T_2$ = initial temperature of water = $25.0^oC$

Now put all the given values in the above formula, we get

$150g\times c_1\times (34.3-150.0)^oC=-200g\times 4.184J/g^oC\times (34.3-25.0)^oC$

$c_1=0.449J/g^oC$

Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).

Therefore, the metal used was iron (the specific heat capacity is $0.449J/g^oC$ ).

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4 years ago

The correct name for hno2 is ________. nitrous acid hyponitrous acid nitric acid pernitric acid hydrogen nitrate

balu736 [363]

Nitrous Acid.

Hyponitrous acid: H2N2O2

Nitric acid: HNO3

Pernitric acid: HNO

4 0

3 years ago

One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is Fe2O3 + 3CO → 2Fe + 3CO2 Su

solong [7]

Answer:

$89.55~\%~of~Fe_2O_3~in~the~sample$

Explanation:

The first step in this reaction is the converstion from Kg of $Fe$ to grams of $Fe_2O_3$ .

$1.19x10^3~Kg~Fe~\frac{1000g~Fe}{1~Kg~Fe}~\frac{1~mol~Fe}{55.84~g~Fe}~\frac{1~mol~Fe_2O_3}{2~mol~Fe}~\frac{159.68~g~Fe_2O_3}{1~mol~Fe_2O_3}~\frac{1~Kg~Fe_2O_3}{1000~g~Fe_2O_3}$

$X~=~1.7x10^3~Kg~Fe_2O_3$

Then we can calculate the percentage of $Fe_2O_3$ in the sample:

$\%~Fe_2O_3~=~\frac{1.7x10^3~Kg~Fe_2O_3}{1.9x10^3~Kg~Sample}*100$

$89.55~\%~of~Fe_2O_3~in~the~sample$

3 0

3 years ago