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abruzzese [7]
3 years ago
10

Please actually try when you answer. If you don't know the answer, don't answer and say that, I don't give a flip if you don't k

now the answer. How does a cell at the end of the first phase of the cell cycle differ from a cell at the end of the second phase?
Chemistry
1 answer:
rewona [7]3 years ago
3 0

Answer:

See the answer below

Explanation:

Knowing that the cell cycle has two phases

  • the first phase which is the interphase; and
  • the second phase, the m phase

<em>A cell at the end of interphase would have grown in volume and have a doubled amount of DNA compared to the same cell at the end of m phase. Such a cell would also have more biochemical contents than its counterpart at the end of m phase.</em>

The interphase is divided into;

  • G1 phase,
  • S phase; and
  • G2 phase

A cell at the end of m phase undergoes massive growth and development at the G1 phase, doubles its DNA at the S phase through replication, and synthesizes proteins at the G2 phase before hopping into the m phase again.

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Calculate the OH− concentration after 53 mL of the 0.100 M KOH has been added to 25.0 mL of 0.200 M HBr. Assume additive vol- um
Andre45 [30]

Answer:

\large \boxed{\text{0.0038 mol/L}}

Explanation:

1. Calculate the initial moles of acid and base

\text{moles of acid} = \text{0.0250 L} \times \dfrac{\text{0.200 mol}}{\text{1 L}} = \text{0.005 00 mol}\\\\\text{moles of base} = \text{0.053 L} \times \dfrac{\text{0.100 mol}}{\text{1 L}} = \text{0.0053 mol}

2. Calculate the moles remaining after the reaction

                   OH⁻     +     H₃O⁺ ⟶ 2H₂O

I/mol:      0.0053       0.005 00

C/mol:    -0.00500   -0.005 00

E/mol:      0.0003              0

We have an excess of 0.0003 mol of base.

3. Calculate the concentration of OH⁻

Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L

\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}

8 0
3 years ago
Calculate the volume of 0.100 m hcl required to neutralize 1.00 g of ba(oh)2 (molar mass = 171.3 g/mol).
dedylja [7]

The balanced chemical equation between HCl and Ba(OH)_{2} is:

2HCl (aq) + Ba(OH)_{2}(aq) -->BaCl_{2}(aq) + 2 H_{2}O(l)

Moles of Ba(OH)_{2} = 1.00 g Ba(OH)_{2} * \frac{1 mol Ba(OH)_{2}}{171.3 g Ba(OH)_{2}} = 0.00584 mol Ba(OH)_{2}

Moles of HCl required to neutralize Ba(OH)_{2}:

0.00584 mol Ba(OH)_{2} * \frac{ 2 mol HCl}{1 mol Ba(OH)_{2}} =   0.01168 mol HCl

Calculating the volume of HCl from moles and molarity:

0.01168 mol HCl * \frac{1 L}{0.100 mol} * \frac{1000 mL}{1 L} = 116.8 mL

7 0
3 years ago
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