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Lelu [443]
3 years ago
10

Unlike the cell wall, the cell membrane is…

Chemistry
1 answer:
irakobra [83]3 years ago
5 0
B: produces energy for the cell
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A large weather balloon filled with helium has a volume of 7.00 x 102 L at STP. Which expression should be used to find the mass
balandron [24]

Answer:

mass of Helium = mole of Helium x Molarmass of Helium

Explanation:

Mole of Helium = volume of Helium/ 22.4dm^3/mol

                          = 7.00 x 10^2/22.4

                          = 31.25mol

Mass of Helium = mole of Helium x Molarmass of Helium

                           = 31.25mol x 4g/mol

                           = 125g

7 0
3 years ago
P= f/a solve it for f
Bess [88]

Answer:

Force = Pressure × Area

Explanation:

Easy, just invert the equation. Transpose the force variable over to the left of the equals sign, and transpose the pressure variable back to the right side.

7 0
2 years ago
50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?
yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

           = 25.00 x 0.200

           = 5.00 m-mol

           = 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

                                           = 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

                                                        = 0.167 M

Now the ICE table :

            $NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M)       0.167                   0            0

C (M)         -x                      +x          +x

E (M)      0.167-x                  x           x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$            $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

    = $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

       = $- \log(1.9054 \times 10^{-6} \ M)$

        =5.72

Now, since pH + pOH = 14

           pH = 14.00 - 5.72

                = 8.279

Therefore the ph is 8.279 at the end of the titration.

8 0
3 years ago
Determine the volume in mL of 0.37 M HClO4(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 2
Dimas [21]

Answer:

14.3mL you require to reach the half-equivalence point

Explanation:

A strong acid as HClO₄ reacts with a weak base as CH₃CH₂NH₂, thus:

CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻

As the reaction is 1:1, to reach the equivalence point you require to add the moles of HClO₄ equal to moles CH₃CH₂NH₂ you add originally. Also, half-equivalence point requires to add half-moles of CH₃CH₂NH₂ you add originally.

Initial moles of CH₃CH₂NH₂ are:

20.8mL = 0.0208L × (0.51mol CH₃CH₂NH₂ / 1L) =

0.0106moles CH₃CH₂NH₂

To reach the half-equivalence point you require:

0.0106moles ÷ 2 = 0.005304 moles HClO₄

As concentration of HClO₄ is 0.37M, volume you require to add 0.005304moles is:

0.005304 moles HClO₄ ₓ (1L / 0.37mol) = 0.0143L =

<h3> 14.3mL you require to reach the half-equivalence point</h3>

7 0
3 years ago
What is the difference between Mitosis and Meiosis?
kondor19780726 [428]
Mitosis produces two diploid (2n) somatic cells that are genetically identical to each other and the original parent cell, whereas meiosis produces four haploid (n) gametes that are genetically unique from each other and the original parent (germ) cell.
7 0
2 years ago
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