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natali 33 [55]
3 years ago
13

a manufacturer claims its cleanser work twice as fast as any other . could testt be performed to support the claim?

Chemistry
1 answer:
grin007 [14]3 years ago
4 0
Yes, other cleansers can be tested and compared to the original manufacturers cleanser to see which works faster.
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A student has a piece of aluminum metal what is the most reasonable assumption a student can make about the metal
Gala2k [10]

Answer:

<h2>- It could be stretched into a thin wire.</h2>

Explanation:

As per the question, the most rational claim that the student can make about the aluminum metal is that 'it could be stretched into a thin wire' without breaking which shows its ductility. It is one of the most significant characteristics of a metal. Metals can conduct electricity in any state and not only when melted. Thus, option A is wrong. Options C and D are incorrect as metals neither have the same shape always nor do they break on hitting with a hammer. Therefore, <u>option E</u> is the correct answer.

3 0
3 years ago
Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang
Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

\Delta (H_{f})_{MnO_{2}} = -520.0 KJ/mole

\Delta (H_{f})_{Al_{2}O_{3}} = -1699.8 KJ/mole

The balanced chemical reaction is,

3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)

Formula used :

\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}

\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]

                        = (-3399.6) + (1560)

                        = -1839.6 KJ



5 0
3 years ago
In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
lisabon 2012 [21]

Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0
3 years ago
A waste collection tank can hold 18,754 kg of methanol, which has a density of 0.788 g/cm3. What is the volume of the waste coll
Ivahew [28]
A is correct hope this helps
4 0
3 years ago
Read 2 more answers
Write L values for the following energy levels. What orbital shapes exist in each of
den301095 [7]

Answer:

n   l    m  

����������������������������������

1   0    0 1s 1 2 2

����������������������������������

2   0    0 2s 1 2  

2   1    1,0,-1 2p 3 6 8

����������������������������������

3   0    0 3s 1 2  

3   1    1,0,-1 3p 3 6  

3   2    2,1,0,-1,-2 3d 5 10 18

����������������������������������

4   0    0 4s 1 2  

4   1    1,0,-1 4p 3 6  

4   2    2,1,0,-1,-2 4d 5 10  

4   3    3,2,1,0,-1,-2,-3 4f 7 14 32

Explanation:

4 0
2 years ago
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