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V125BC [204]
3 years ago
12

The mass of a large order of french fries is about 170 G what is its approximate weight in pounds

Chemistry
1 answer:
kap26 [50]3 years ago
8 0

Answer:

             0.374 Pound Approximately

Solution:

Gram and Pound are related as,

1 Gram  =  0.0022 Pound (approximately)

As we are given with 170 g of French Fries, so it can be converted into pounds as,

          1 Gram  =  0.0022 Pound

So,

          170 Grams  =  X Pounds

Solving for X,

                      X =  (170 Gram × 0.0022 Pounds) ÷ 1 Gram

                      X =  0.374 Pound Approximately

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Vaselesa [24]
The chemical reaction is given as:

<span>C8H10+3O2 = C8H6O4+2H2O

We are given the initial amount of para-xylene for the reaction. We use this amount and the relation of substances from the reaction to obtain the amount of the acid produced.

204 g ( 1 mol / 106.18 g ) ( 1 mol C8H6O4 / 1 mol C8H10 ) ( 166.14 g / mol ) = 319.20 g terephthalic acid</span>
4 0
3 years ago
What is the concentration of H+ in 0.0025 M HClO4? What is the pH of the solution? What is the OH− concentration in the solution
alexandr402 [8]

Answer:

A. The concentration of H+ is 0.0025 M

B. The pH is 2.6

C. The concentration of OH- is 3.98x10^-12 M

Explanation:

We'll begin by writing the balanced

dissociation equation of HClO4. This is illustrated below:

HClO4 —> H+ + ClO4-

A. Determination of the concentration of H+ in 0.0025 M HClO4. This is illustrated below:

From the balanced equation above,

1 mole of HClO4 produced 1 mole of H+.

Therefore, 0.0025 M of HClO4 will also produce 0.0025 M of H+.

The concentration of H+ is 0.0025 M

B. Determination of the pH.

The pH of the solution can be obtained as follow:

The concentration of H+, [H+]

= 0.0025 M

pH =?

pH = - log [H+]

pH = - log 0.0025

pH = 2.6

C. Determination of the concentration of OH-

To obtain the concentration of OH-, we must first calculate the pOH of the solution. This is illustrated below:

pH + pOH = 14

pH = 2.6

pOH =?

pH + pOH = 14

2.6 + pOH = 14

Collect like terms

pOH = 14 - 2.6

pOH = 11.4

Now, we can calculate the concentration of the OH- as follow:

pOH = - Log [OH-]

pOH = 11.4

11.4 = - Log [OH-]

- 11.4 = log [OH-]

[OH-] = anti log (- 11.4)

[OH-] = 3.98x10^-12 M

8 0
3 years ago
Lithium nitride reacts with water to produce ammonia and lithium hydroxide according to the equation, Li3N(s) + 3H2O (L) --&gt;
Gemiola [76]

Answer:

0.480 grams

Explanation:

Li₃N(s) + 3D₂O (L) --------------------------> ND₃(g) + 3LiOD (aq)

1             :  3                                            : 1           :  3

Number of moles (n) = Mass in gram/ Molar Mass

Mass of ND₃ = 160 mg

                     = 0.16 g

Molar mass of ND₃=  [14 + (3 x 2.014 )]

                               =    14 + 6.042

                               =  20.042 g/mol

Number of moles of   ND₃  =  0.16/20.042

                                             =  0.007983 moles

From the reaction equation, the mole ratio between   Heavy water (D₂O ) and  ND₃ is  3: 1.

This implies that the number of moles of   Heavy water (D₂O )  required

= 3  x 0.007983 moles

=  0.023949 moles

Molar mass of  Heavy water (D₂O )=  [(2.014 x 2) + 16]

                                                           =  20.028 g/mol

Mass in grams of Heavy water (D₂O )= Number of moles  x Molar mass

                                                            =   0.023949   x  20.028

                                                            =  0.4797 grams

                                                            ≈ 0.480 grams

3 0
3 years ago
2Al + 3H2SO4 -&gt; Al2(SO4)3 + 3H2How many grams of aluminum sulfate would be formed if 250g H2SO4 completely reacted with alumi
bazaltina [42]

Answer:

290.82g

Explanation:

The equation for the reaction is given below:

2Al + 3H2SO4 -> Al2(SO4)3 + 3H2 now, let us obtain the masses of H2SO4 and Al2(SO4)3 from the balanced equation. This is illustrated below:

Molar Mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 +64 = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar Mass of Al2(SO4)3 = (2x27) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 54 + 288 = 342g

Now, we can obtain the mass of aluminium sulphate formed by doing the following:

From the equation above:

294g of H2SO4 produced 342g of Al2(SO4)3.

Therefore, 250g of H2SO4 will produce = (250 x 342)/294 = 290.82g of Al(SO4)3

Therefore, 290.82g of aluminium sulphate (Al(SO4)3) is formed.

7 0
3 years ago
IF YOU ANSWER FIRST, I'll give you the crown hehe
WINSTONCH [101]

Answer:

True

Explanation:

8 0
3 years ago
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