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2 years ago

Please answer this question!

Chemistry

1 answer:

Hunter-Best [27]2 years ago

3 0

1,2-dimbromopropane would be something like C which is the second best answer.

B would be correct if it had another carbon.

A is like C but incorrect for the same reason. 1,2-dibromoethane is what A is.

The answer has to be D where the bromines are right across from each other.

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A scientist adds a liquid to a base. As it is added the substance turns reddish pink the liquid is most likely?

Natalka [10]

Answer:

what we need to answer here?

6 0

3 years ago

Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +

qwelly [4]

Answer: The $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)$ $\Delta H_1=-668.5kJ$

(2) $XCO_3(s)\rightarrow XO(s)+CO_2$ $\Delta H_2=+384.3kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

7 0

3 years ago

A slurry of flakes soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt% oil and

lubasha [3.4K]

Answer:

the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.

Explanation:

Let state the given parameters;

Let A= solvent (hexane)

B= solid(inert soiid)

C= solvent(oil)

$F_{solution}$ = mass of solvent + mass of oil (i.e A+C)

Feed Phase:

Total feed (i.e slurry of flakes soybeans)= 100kg

B= mass of solid =75 kg

F= mass of solvent + mass of oil (i.e A+C)

= 25kg

Mass ratio of oil to solution $Y_{F}$ = $\frac{Mass C}{Mass (A+C)}$

mass of oil (C) =25 × 0.1 wt = 2.5kg

mass of hexane in feed = 25 × 0.9 =22.5kg + 2.5 =25kg

therefore $Y_{F}$ = $\frac{2.5}{25}$

= 0.1

mass ratio of solid to solution $Y_{A}$ = $\frac{Mass A}{Mass (A+C)}=[tex]\frac{75}{25}$

Solvent Phase:

C= Mass of oil= 0(kg)

A= Mass of hexane = 100kg

mass of solutions = A+C = 0+100kg

solvent= 100kg

Underflow:

underflow = L₁ = (unknown) ???

L₁ = E₁ + B

the value of N for the outlet and underflow is 1.5 kg

i.e N₁ = $\frac{mass B}{mass(A+C)}$

solution in underflow E₁ = Mass (A+C)

Overflow:

Overflow = V₁ = (unknown) ???

solution in overflow V₁ = Mass (A+C)

This is because, B = 0 in overflow

Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).

solid in feed = solid in underflow = 75

75= E₁ × N₁

75 = E₁ × 1.5

E₁ = 50kg

Underflow L₁ = E₁ × B

= 50 + 75

=125kg

The Overall Balance: Feed + Solvent = underflow + overflow

100 + 100 = 125 + V₁

V₁ = 75kg

5 0

3 years ago

How many grams of neutral salt will be obtained in the reaction of calcium oxide with 200 cm 3 of phosphoric acid solution whose

katen-ka-za [31]

Answer:

9.3 g of Ca3(PO4)2

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

Next, we shall determine the number of mole of H3PO4 present in 200 cm³ of 0.3 mol/dm³ phosphoric acid (H3PO4) solution. This can be obtained as follow:

Molarity of H3PO4 = 0.3 mol/dm³

Volume = 200 cm³ = 200 cm³/1000 = 0.2 dm³

Mole of H3PO4 =?

Molarity = mole /Volume

0.3 = mole of H3PO4 /0.2

Cross multiply

Mole of H3PO4 = 0.3 × 0.2

Mole of H3PO4 = 0.06 mole

Next, we shall determine the number of mole of the salt, Ca3(PO4)2, obtained from the reaction. This can be obtained as shown below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

From the balanced equation above,

2 moles of H3PO4 reacted to produced 1 mole of Ca3(PO4)2.

Therefore, 0.06 moles of H3PO4 will react to produce = (0.06 × 1)/2 = 0.03 mole of Ca3(PO4)2.

Thus, 0.03 mole of Ca3(PO4)2 is produced from the reaction.

Finally, we shall determine the mass of Ca3(PO4)2 produced as follow:

Mole of Ca3(PO4)2 = 0.03 mole

Molar mass of Ca3(PO4)2 = (40×3) + 2[31 + (16×4)]

= 120 + 2[31 + 64]

= 120 + 2[95]

= 120 + 190

= 310 g/mol

Mass of Ca3(PO4)2 =?

Mole = mass /Molar mass

0.03 = mass of Ca3(PO4)2 / 310

Cross multiply

Mass of Ca3(PO4)2 = 0.03 × 310

Mass of Ca3(PO4)2 = 9.3 g

Thus, 9.3 g of Ca3(PO4)2 was obtained from the reaction.

6 0

2 years ago

HELP PLZ

Vinvika [58]

Answer:

1). 19.9

2). 0

8 0

2 years ago