What is a fusion reaction

How many kilojoules are released when 8.2 g of water condenses at 100 °c and cools to 15 °c?

MrRa [10]

Answer:- 2.92 kJ of heat is released.

Solution:- We have water at 100 degree C and it's going to be cool to 15 degree C.

So, change in temperature, $\Delta T$ = 15 - 100 = -85 degree C

mass of water, m = 8.2 g

specific heat of water, c = $4.184\frac{J}{^0C.g}$

The equation used for solving this type of problems is:

$q=mc\Delta T$

Let's plug in the values in the equation and solve it for q which is the heat energy:

q = (8.2)(4.184)(-85)

q = -2916.248 J

They want answer in kJ. So, let's convert J to kJ and for this we divide by 1000.

$q=-2916.248J(\frac{1kJ}{1000J})$

q = -2.92 kJ

Negative sign indicates the heat is released. So, in the above process of coiling of water, 2.92 kJ of heat is released.

3 0

3 years ago

Homeostasis means maintaining

Zielflug [23.3K]

Answer:

Homeostasis is the tendency to resist change in order to maintain a stable, relatively constant internal environment. Homeostasis typically involves negative feedback loops that counteract changes of various properties from their target values, known as set points.

Explanation:

So D. Glad to help! :D

4 0

3 years ago

Read 2 more answers

Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.

Moles of $H_2O$ reacted = 1.05 mol $Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}$

= 6.31 mol $H_2O$

The mass of $H_2O$ is,

$m_{H_{2} O}$ = 6.31 mol $\times$ 18.02g/ mol

= 114g

On subtracting, the mass of $H_2O$ reacted from the given mass of $H_2O$ is,

$m_{H_2O}$ = (131-114)g

= 17g

Hence, the excess remaining reactant is 17g

To learn more about Chemical Reaction

brainly.com/question/11231920

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8 0

2 years ago

Sediments are almost always deposited in flat layers. Does it appear that forces in the Earth affected the rock layers in this r

laiz [17]

I think so but what region do you live in

4 0

3 years ago

How can valid conclusions from the growth rate calculations be applied in a biotechnological or biomedical context ?

Igoryamba

Growth rate calculations and their conclusions can be used in biomedical contexts when organisms are being grown for the synthesis of compounds or other purposes.

7 0

3 years ago