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Bezzdna [24]
3 years ago
13

I will give brainiest answer to whoever helps me, also this problem is worth 20 points

Chemistry
1 answer:
Svetlanka [38]3 years ago
8 0
It is absorbing because it is going through endothermic process (absorbing heat) and the change in color means it is undergoing a chemical reaction
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5. Balance the equation and state the limiting reagent in the following reaction:
mr Goodwill [35]

Answer:

5. 2 Al  3 Cl2 and 2 AlCl3

Cl2 is limiting

(It has the highest coefficients, but that's not always true. If they give you the number of grams for each make sure to work the math out)

6. C

(As the pressure increases there is less volume for the molecules to move around, therefore the molecules will be closer together, allowing for more collisions.)

7.  b

(if the temperature is increases so will the rate of the reaction. The increase in temperature will allow for the molecules to move at a faster rate, allowing for more collisions (reactions) between them.

8. d

(dynamic equilibrium is the state at which the reaction is consistently moving, there fore the amounts should remain the same)

9. A

(a catalyst will lower the the pathway of activation energy because it's meant to speed up a reaction)

bro youre on your own after this, im tired

8 0
3 years ago
Consider this reaction: HCO3− + H2S → H2CO3 + HS− Which is the Bronsted-Lowry base? H2S HCO3- HS– H2CO3
german

Answer:

hco3

Explanation: bc i said so

3 0
3 years ago
qué papel desempeña el narrador en la historia circula con el negro el párrafo en el que investiga el narrador e indica sin larg
liubo4ka [24]
No mms wey qué pasa vato
6 0
3 years ago
Assume that an exhaled breath of air consists of 74.8% N2, 15.3% O2, 3.7% CO2, and 6.2% water vapour.
Goryan [66]

Answer:

a) PN₂ = 0.733 atm

PO₂ = 0.150 atm

PCO₂ = 0.036 atm

Pwater = 0.061 atm

b) 6.44x10⁻⁴ mol

c) 0.02 g

Explanation:

a) By the Dalton's Law, in a gas mixture, the total pressure is the sum of the partial pressures, and the partial pressure is the molar fraction of the gas multiplied by the total pressure.

PN₂ = 0.748*0.980 =0.733 atm

PO₂ = 0.153*0.980 = 0.150 atm

PCO₂ = 0.037*0.980 = 0.036 atm

Pwater = 0.062*0.980 = 0.061 atm

b) The number of moles of CO₂ can be calculated by the ideal gas law:

PV = nRT, where P is the pressure, V is the volume (0.455 L), n is the number of moles, R is the gas constant (0.082 atm.L/mol.K), and T is the temperature (37°C + 273 = 310 K).

0.036*0.455 = 0.082*310*n

25.42n = 0.01638

n = 6.44x10⁻⁴ mol

c) For the combustion reaction of glucose:

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

So, the stoichiometry is:

1 mol of glucose ------- 6 moles of CO₂

x ------- 6.44x10⁻⁴ mol of CO₂

By a simple direct three rule:

6x = 6.44x10⁻⁴

x = 1.073x10⁻⁴mol of glucose

Glucose has a molar mass equal to 180 g/mol, and its mass is the molar mass multiplied by the number of moles:

m = 180x1.073x10⁻⁴

m = 0.02 g

8 0
3 years ago
g An oxidized silicon (111) wafer has an initial field oxide thickness of d0. Wet oxidation at 950 °C is then used to grow a thi
77julia77 [94]

Answer:

Explanation:

From the information given:

oxidation of oxidized solution takes place at 950° C

For wet oxidation:

The linear and parabolic coefficient can be computed as:

\dfrac{B}{B/A} = D_o \ exp \Big [\dfrac{-\varepsilon a}{k_BT} \Big]

Using D_o and E_a values obtained from the graph:

Thus;

\dfrac{B}{A} = 1.63 \times 10^8 exp \Big [ \dfrac{-2.05}{8.617 \times 10^{_-5}\times 1173}\Big] \\ \\ = 0.2535 \ \ \mu m/hr

B= 386 \  exp \Big [-\dfrac{0.78}{8.617 \times 10^{-3} \times 1173} \Big] \\ \\  = 0.1719 \ \mu m^2/hr

So, the initial time required to grow oxidation is expressed as:

t_{ox} = \dfrac{x}{B/A}+ \dfrac{x^2}{B} - t_o (initial)

where; \\ \\ t_{ox} = 2 \ hrs;\\ \\ x = 0.5 \\ \\ B/A = 0.2535 \\ \\  B = 0.1719

∴

2= \dfrac{0.5}{0.2535}+ \dfrac{0.5^2}{0.1719} - t_o (initial)

2 = 3.4267 - t_o (initial) \\ \\ t_o(initial) = 3.4267 - 2  \\ \\ t_o(initial) = 1.4267 \ hr

NOW;

1.4267 = \dfrac{d_o}{0.2535} + \dfrac{d_o^2}{0.1719} \\ \\  1.4267 = 3.9448 \ d_o + 5.8173 \ d_o^2 \\ \\ d_o^2 + 0.6781 \ d_o = 0.2453

d_o = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}

d_o = \dfrac{-(0.6781) \pm \sqrt{(0.6781)^2-4(1)(-0.245)}}{2(10)}

d_o = \dfrac{-(0.6781) \pm \sqrt{0.45981961+0.98}}{20}

d_o = \dfrac{-(0.6781) \pm \sqrt{1.43981961}}{20}

d_o = \dfrac{-(0.6781) + \sqrt{1.43981961}}{20} \ OR \   \dfrac{-(0.6781) - \sqrt{1.43981961}}{20}

d_o =0.02609 \ OR \   -0.0939

Thus; since we will consider the positive sign, the initial thickness d_o is ;

≅ 0.261 μm

3 0
3 years ago
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