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3 years ago

What is the method use to separate sand and water

Chemistry

1 answer:

Slav-nsk [51]3 years ago

7 0

Sand and water can be separated by any of the following methods:

1. Sedimentation and decantation: This method involves the mixture being kept undisturbed for some time. After some time, sand being heavier and insoluble in water, settles down at the bottom of container. Now, water is poured into another container to separate it from sand.

2. Filtration: This method involves the mixture being passed through a filter paper (a filter with very fine pores). Sand particles being larger in size are retained by the filter paper and get separated from water.

I hope this helps! :D

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Consider the following equilibrium:

irga5000 [103]

Consider the following equilibrium:

4NH₃ (g) + 3O₂ (g) ⇄ 2N₂ (g) + 6H₂O (g) + 1531 kJ

The given statement is True, because

According to Le Ch's Principle:

Systems that have attained the state of chemical equilibrium will tend to maintain their equilibrium state.

External factors such as the addition of products and reactants result in the disruption of the equilibrium state.

we expect the system to shift to the direction that offsets the change in concentration.

This results in the state of chemical equilibrium to be reestablished.

Hence, The statement is true,

The addition of more ammonia (a reactant) would offset the state of equilibrium.
To restore chemical equilibrium, the system must consume the excess reactants to form more products.
A shift to favor the products side occurs.

To learn more about equilibrium here

brainly.com/question/3920294

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8 0

2 years ago

W + BgCz2 --> WCz + Bg Balance this equation

yanalaym [24]

Answer: W + BgCz2 arrow WCz + Bg

2 W + BgCz2 arrow 2 WCz + Bg

Explanation:

Cz has 2 so you balcne the other side of WCz.

Since you Balcanes the Cz you changed the W and you Balcanes the other W on the left side.

3 0

3 years ago

4). One mole of monoclinic sulfur at 25C was placed in a constant-pressure calorimeter whose heat capacity (C) was 1620 J/K. T

andre [41]

Answer: The enthalpy change of the reaction is -243 J/mol

Explanation:

The heat released by the reaction is absorbed by the calorimeter and the solution.

The chemical equation used to calculate the heat released follows:

$q=c\times \Delta T$

where,

c = heat capacity of calorimeter = 1620 J/K

$\Delta T$ = change in temperature = $0.150^oC=0.150K$ (Change remains same)

Putting values in above equation, we get:

$q=1620J/K\times 0.15K=243J$

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

For the given chemical reaction:

$S\text{ (monoclinic)}\rightarrow S\text{ (orthorhombic)}$

We are given:

Moles of monoclinic sulfur = 1 mole

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = -243 J

n = number of moles = 1 mole

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-243J}{1mol}=-243J/mol$

Hence, the enthalpy change of the reaction is -243 J/mol

8 0

4 years ago

An acid with molar mass 84.48 g/mol is titrated with 0.650 M KOH. What volume of KOH solution is needed to titrate 1.70 grams of

RSB [31]

Answer:

$V=0.0310L=3.10mL$

Explanation:

Hello.

In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

$n_{acid}=n_{KOH}$

Thus, since we are given 1.70 g of the acid, we compute the moles of acid that were titrated:

$n_{acid}=1.70g*\frac{1mol}{84.48g}=0.0201mol$

Which equal the moles of KOH. In such a way, since the molarity is defined as moles over liters (M=n/V), the liters are moles over molarity (V=n/M), thus, the resulting volume is:

$V=\frac{0.0201mol}{0.650mol/L}\\\\V=0.0310L=3.10mL$

Best regards!

7 0

3 years ago

How many liters of oxygen gas, at standard

Karo-lina-s [1.5K]

Answer:

Explanation:

For the balanced reaction:

4Fe(s) + 3O₂(g) → 2Fe₂O₃(s).

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

Using cross multiplication:

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

Finally, we can get the volume of oxygen using the information:

It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).

Using cross multiplication:

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

5 0

4 years ago

What is the method use to separate sand and water​

What is the method use to separate sand and water