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Yuki888 [10]
3 years ago
7

The Faruq family uses 548 liters of water per day.

Chemistry
1 answer:
Radda [10]3 years ago
4 0

Answer:

11508 liters

Explanation:

The Faruq family uses 548 liters of water per day.  The Chammas family uses 3 times as much water  per day. How much water does the Chammas family use per week?

Solution:

The amount of water used by the Chammas family is three times that of the Faruq family per day, hence:

Amount of water used by Chammas family in a day = 3 * water used by Faruq family in a day

Amount of water used by Chammas family in a day = 3 * 548 = 1644 liters

The amount of water used by the Chammas family in a week = 1644 liters per day * 7 days in a week = 11508 liters

Therefore 11508 liters of water was used by the Chammas family in a week

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Consider these equations: 2S(s)+3O2(g)→2SO3(g) , ΔH=−792kJ 2S(s)+2O2(g)→2SO2(g) , ΔH=−594kJ 2SO2(g)+O2(g)→2SO3(g) , ΔH=? What is
Elenna [48]

Hess law can help us to obtain the enthalpy of a series of reactions by summation. The missing enthalpy of this reaction is −198kJ.

<h3>What is  Hess law?</h3>

According to the Hess law of constant heat summation, the enthalpy of a sequence of reactions can be obtained as the sum of the enthalpies of all the reactions.

Now looking at all the reavtion written here in order to obtain the missing enthalpy, we conclude that the missing enthalpy is −198kJ.

Learn more about enthalpy: brainly.com/question/3393755

6 0
2 years ago
21+ Cl2 → 12+201 -
Alina [70]

Answer:

\rm 2\; I^{-} + Cl_2 \to I_2 + 2 \; Cl^{-}.

Start color: yellowish-green.

End color: dark purple.

Assumption: no other ion in the solution is colored.

Explanation:

In this reaction, chlorine gas \rm Cl_2 oxidizes iodine ions \rm I^{-} to elemental iodide \rm I_2. At the same time, the chlorine atoms are converted to chloride ions \rm Cl^{-}.

Fluorine, chlorine, bromine, and iodine are all halogens. They are all found in the 17th column of the periodic table from the left. One similarity is that their anions are not colored. However, their elemental forms are typically colored. Besides, moving down the halogen column, the color becomes darker for each element.

Among the reactants of this reaction, \rm I^{-} is colorless. If there's no other colored ion, only the yellowish-green hue of \rm Cl_2 would be visible. Hence the initial color of the reaction would be the yellowish-green color of \rm Cl_2.

Similarly, among the products of this reaction, \rm Cl^{-} is colorless. If there's no other colored ion, only the dark purple hue of \rm I_2 would be visible. Hence the initial color of the reaction would be the dark purple color of \rm I_2.

5 0
2 years ago
Silicon is prepared by the reduction of K₂SiF6 with Al. Write the equation for this reaction. (Hint: Can F⁻ be oxidized in this
BlackZzzverrR [31]

4Al + 3K2SiF6 = 6KF + 3Si + 4AIF3 is the reaction for preparation of silicon by the reduction of K₂SiF6 with Al.

AlF3xH2O-based inorganic compounds are referred to as aluminium fluoride. They are all solids without colour. Aluminium fluoride is a crystalline (sand-like), odourless, white, or colourless powder. In addition to being used to make aluminium, it also functions as a flux in welding processes and in ceramic glazes and enamels.

Silicon (Si) is created by reducing potassium silicofluoride with aluminium as the reducing agent (K2SIF6). While K2SiF6 is reduced to Si in this equation, aluminium is oxidised to aluminium fluoride. As a result, the balanced equation describing aluminum's reduction of K2SiF6 to silicon non-metal is as follows: 4Al + 3K2SiF6 = 6KF + 3Si + 4AIF3

Learn more about aluminium fluoride here:

brainly.com/question/17131529

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3 0
10 months ago
The second-order diffraction for a gold crystal is at an angle of 22.20o for X-rays of 154 pm. What is the spacing between the c
Alenkinab [10]

<u>Answer:</u> The spacing between the crystal planes is 4.07\times 10^{-10}m

<u>Explanation:</u>

To calculate the spacing between the crystal planes, we use the equation given by Bragg, which is:

n\lambda =2d\sin \theta

where,

n = order of diffraction = 2

\lambda = wavelength of the light = 154pm=1.54\times 10^{-10}m     (Conversion factor:  1m=10^{12}pm )

d = spacing between the crystal planes = ?

\theta = angle of diffraction = 22.20°

Putting values in above equation, we get:

2\times 1.54\times 10^{-10}=2d\sin (22.20)\\\\d=\frac{2\times 1.54\times 10^{-10}}{2\times \sin (22.20)}\\\\d=4.07\times 10^{-10}m

Hence, the spacing between the crystal planes is 4.07\times 10^{-10}m

4 0
2 years ago
What type of property is mass
BabaBlast [244]

Answer:

I'm pretty sure its solid

6 0
3 years ago
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