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Ann [662]
3 years ago
9

In a titration of 0.35 M HCl and 0.35 M NaOH, how much NaOH should be added to 45.0 ml of HCl to completely neutralize the acid?

Chemistry
1 answer:
Bezzdna [24]3 years ago
6 0
It would be the same amount. So, 45 ml of NaOH is required to be added to the 45 ml of HCI to neutralize the acid fully. Here is a brief calculation:

Firstly, here is your formula: M(HCI) x V(HCI) = M(NaOh) x V(NaOH) 
With the values put in: 0.35 x 45 = 0.35 x V(NaOH) 
= 45 ml. 
There is 45 ml of V(NaOH)

Let me know if you need anything else. :)

           - Dotz
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A sample of nitrogen gas has a volume of 478 cm3 and a pressure of 104.1 kPa. What volume would the gas occupy at 88.2 kPa if th
nadezda [96]

<span>1.    </span>To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

V2 = P1 x V1 / P2

 V2 = 104.1 x 478 / 88.2

<span> V2 =564.17 cm^3</span>

6 0
3 years ago
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4 Al + 3 O2 --&gt; 2 Al2O3
SIZIF [17.4K]

Answer:

The answer to your question is 7.4 moles of Aluminum

Explanation:

Data

moles of Al = ?

moles of Al₂O₃ = 3.7

Balanced chemical reaction

                4 Al  +  3 O₂  ⇒  2 Al₂O₃

To solve this problem use proportions and cross multiplication. Use the coefficients of the balanced chemical equation.

                4 moles of Aluminum ----------------- 2 moles of Al₂O₃

                 x                                  ----------------- 3.7 moles of Al₂O₃

                            x = (3.7 x 4) / 2

                            x = 14.8 / 2

                            x = 7.4 moles of Aluminum

5 0
3 years ago
How many moles are in 39.5 grams of Lithium?
Blizzard [7]

Answer:

185.05 g.

Explanation

Firstly, It is considered as a stichiometry problem.

From the balanced equation: 2LiCl → 2Li + Cl₂

It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.

We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.

n = (30.3 g) / (6.941 g/mole) = 4.365 moles.

Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.

Using cross multiplication:

2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.

??? moles of LiCl → 4.365  moles of Li.

The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.

Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).

Molar mass of LiCl = 42.394 g/mole.

mass = n x molar mass = (4.365 x 42.394) = 185.05 g.

7 0
3 years ago
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Answer:

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Answer:

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Nuclear membrane begins to re-form. - this is false. This happens during telophase.

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