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zheka24 [161]
3 years ago
11

The structure is primarily double helixes

Chemistry
1 answer:
zheka24 [161]3 years ago
8 0
Are you referring to DNA? 
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Explain how engineers and scientists may work together to produce a protein in the lab.
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The scientists would do biological studies of how the protein breakdown and combines with the muscles the engineers with then create a delivery system to get the protein to the muscle quicker and more effectively
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Give the expression for the solubility product constant for baf2.
svetoff [14.1K]

<span>Solubility product constant (Ksp) is </span>applied to the saturated ionic solutions<span> which are in equilibrium with its solid form. The solid is partially dissociated into its ions.</span><span>

For the BaF, the dissociation as follows;
BaF</span>₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)

<span>
Hence,
        Ksp = [Ba</span>²⁺(aq)] [F⁻(aq)]²

3 0
3 years ago
Given their location on the periodic table, identify the ionic charge for each element and predict the compound formed by the ba
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Barium has a 2+ charge as it is in group 2 in the periodic table and so it has two electrons in its outer shell and chloride has a -1 charge on its chloride ion. So we will need two of the chloride ions as we have a 2+ charge to match the amount of charge on one barium ion- forming barium ion

BaCI2
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3 years ago
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Use the molar solubility 3.27×10−11m in pure water to calculate ksp for nis.
Ne4ueva [31]

Answer:

Ksp = 1.07x10⁻²¹

Explanation:

Molar solubility is defined as moles of solute can be dissolved in 1L.

Ksp for NiS is defined as:

NiS(s) ⇄ Ni²⁺(aq) + S²⁻(aq)

Ksp = [Ni²⁺] [S²⁻]

As molar solubility is 3.27x10⁻¹¹M, concentration of [Ni²⁺] and [S²⁻] is 3.27x10⁻¹¹M for both.

Replacing:

Ksp = [3.27x10⁻¹¹M] [3.27x10⁻¹¹M]

<em>Ksp = 1.07x10⁻²¹</em>

<em></em>

4 0
3 years ago
If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was
Andru [333]

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

n_{acid}=n_{base}=n_{salt}

Whereas the moles of the salt are computed as shown below:

n_{salt}=0.021L*0.68mol/L=0.01428mol

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:

[salt]=0.01428mol/0.0276L=0.517M

Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

C_6H_5NH_3^++H_2O\rightleftharpoons  C_6H_5NH_2+H_3O^+

Whose equilibrium expression is:

Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}

Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

2.326x10^{-5}=\frac{x^2}{0.517M}

Whereas x is:

x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

pH=-log(3.47x10^{-3})\\\\pH=2.46

Regards!

6 0
3 years ago
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