Answer:
12.29 M
Explanation:
- The reaction that takes place is:
H₂SO₄ + 2NaOH → 2Na⁺ + SO₄⁻² + 2H₂O
- Now let's calculate the <u>moles of H₂SO₄ that were titrated</u>:
= 0.01229 mol H₂SO₄.
- Thus, the <u>concentration of the diluted solution is</u>:
0.01229 mol H₂SO₄ / 0.010 L = 1.229 M
- Finally, the <u>concentration of the original acid solution is:</u>
= 12.29 M
There are:
3.41 moles of C
4.54 moles of H
3.40 moles of O.
Why?
To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.
We have that:

To know the percent of each element, we need to to the following:

So, we know that for the 100 grams of the compound, there are:
40.92 grams of C
4.58 grams of H
54.49 grams of O
We know the molecular masses of each element:

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.
Have a nice day!
Given :
Moles of Na : 1.06
Moles of C : 0.528
Moles of O : 1.59
To Find :
The empirical formula of the compound.
Solution :
Dividing moles of each atom with the smallest one i.e 0.528 .
So,
Na : 1.06/0.528 = 2.007 ≈ 2
C : 0.528/0.528 = 1
O : 1.59/0.528 = 3.011 ≈ 3
Rounding all them to nearest integer, we will get the number of each atom in the empirical formula.
So, empirical formula is
.
Hence, this is the required solution.
6.28×1013+7.30×1011 this =13741.94
Answer:
I can give you the definition ... That might help cause I honestly don't kno the answer either ;-;
Explanation:
When used as a diacritic mark, the term dot is usually reserved for the interpunct, or to the glyphs 'combining dot above' and 'combining dot below' which may be combined with some letters of the extended Latin alphabets in use in Central European languages and Vietnamese.
Here is an example:
The dot product between a unit vector and itself is also simple to compute. In this case, the angle is zero and cosθ=1. Given that the vectors are all of length one, the dot products are i⋅i=j⋅j=k⋅k=1.