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3 years ago

Help me with this question?!!

Chemistry

1 answer:

Ne4ueva [31]3 years ago

5 0

Answer:

Glow sticks and match would be light emission, slime would be preciptate, and cookies would be gas.

Explanation:

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Zn + 2HCI --> ZnCl2 + H2

Elena-2011 [213]

Moles= mass\ relative formula mass(Ar)
moles of zinc= 7.9/30= 0.263
so we have 0.263 moles of zinc, and you need twice the amount of chlorine so therefore 0.526moles of chlorine= 0.526x 17=8.942g of chlorine
i cba to work the rest out but the most reasonable answer is 0.24 mol however if you need to use working outs, use the formula i provided earlier

8 0

3 years ago

A mixture containing 20 mole % butane, 35 mole % pentane and rest

notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

$Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB$

Whereas $y$ accounts for the fractions at the outlet distillate and $x$ for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

$0.9=\frac{y_bD}{z_bF}$

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

$y_bD=0.9*z_bF=0.9*0.2*100mol=18mol$

The total distillate flow:

$y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol$

And the total bottoms flow:

$F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol$

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

$x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025$

Then, the molar fraction of pentane and hexane:

$x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422$

$x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553$

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0

3 years ago

Read 2 more answers

1. Which of the following is a correctly written thermochemical equation?

WITCHER [35]

Explanation:

1. Thermochemical equation is balance stoichiometric chemical equation written with the phases of the reactants and products in the brackets along with the enthalpy change of the reaction.

The given correct thermochemical reactions are:

$Fe(s)+O_2(g)\rightarrow Fe_2O_3(s),\Delta H = 3,926 kJ
$

$C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2 (g)+4H_2O(l),\Delta H= 2,220 kJ/mol$

2. Phase change affect the value of the enthalpy change of the thermochemical equation. This is because change in phase is accompanied by change in energy. For example:

$H_2O(s)\rightarrow H_2O(g),\Delta H_{s}=51.1 kJ/mol$

$H_2O(l)\rightarrow H_2O(g),\Delta H_{v}=40.65 kJ/mol$

In both reaction phase of water is changing with change in energy of enthalpy of reaction.

7 0

3 years ago

Read 2 more answers

Balance the following equation. Then, given the moles of reactant or product below, determine the corresponding amount in moles

Mrrafil [7]

Answer:

Option a → 4 mol NH₃

Explanation:

This the unbalanced reaction

NH₃ + O₂ ⟶ N₂ + H₂O

The balanced reaction:

4NH₃ + 3O₂ → 2N₂ + 6H₂O

4 mol of ammonia

3 mol of oxygen

2 mol of nitrogen

6 mol of water

6 0

3 years ago

In Rutherford's Gold foil experiment, were the alpha particles directed to different areas of the gold foil or only the same spo

MrRissso [65]

Answer:

See explanation

Explanation:

In the Rutherford experiment, alpha particles were directed at the same spot on a thin gold foil.

As the alpha particles hit the foil, most of the alpha particles went through the foil. In Rutherford's interpretation, most of the particles went through because the atom consisted largely of empty space.

However, some of the alpha particles were deflected through large angles, in Rutherford's interpretation, the deflected alpha particles had hit the dense positive core of the atom which he called the nucleus.

This accounted for their scattering through large angles throughout the foil in all directions.

8 0

3 years ago