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3 years ago

During transcription, RNA polymerase encounters the sequence of DNA

Chemistry

2 answers:

mamaluj [8]3 years ago

8 0

Answer:

C. TCG CGA

Explanation:

A is replaced by T, C is replaced by G, and vice versa

Here is an easy way to memorize it:

Apples on Trees and Cars in Garage

SVEN [57.7K]3 years ago

3 0

Answer:

it 123456789 •-• ™ I h--e yu hehehe

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Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

8 0

4 years ago

Two hydraulic cylinders are connected. If the diameter of one piston is twice the other, the how does the pressure experienced b

Aneli [31]

Answer:

They experience the same pressure

Explanation:

To answer this question, we recall Pascal's, Law Pascal's law states that an increase in pressure at a point in a confined cylinder containing a fluid, there is also an equal increase at all other points in that cylinder.

According to Pascal's law the pressure if the pressure expereienced by the larger diameter piston increases, the pressure experienced by the smaller diameter piston also increases by the same amount

However considering that pressure = Force/area F1/A1 =F2/A2

thus where A1 = πD²÷4 and A2 = πD²÷ 16 we have

we have F1×4/πD² = F2×16/πD² or F1 = 4× F2

They experience the same pressure but the larger cylinder delivers four times the force transmitted from he outside to the smaller cylinder

7 0

3 years ago

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Mama L [17]

I can’t read ? This what does it say

7 0

3 years ago

Would precipitation occur when 500 mL of a 0.02M solution of AgNO3 is mixed with 500 mL of a 0.001M solution of NaCl? Show your

Oksanka [162]

We know,
AgNO3 + NaCl ⇒ NaNO3 + AgCl(s)
The moles of Na+ present:
0.5 L * 0.001 mol/L
= 5 x 10⁻⁴ mol
Moles of Ag+ present:
0.5 * 0.02
= 0.01 mol
The limiting reactant is Na
Therefore, the moles of Ag reacted:
5 x 10⁻⁴
AgCl is insoluble in water; therefore, the AgCl formed will precipitate

7 0

3 years ago

2. Suppose that 21.37 mL of NaOH is needed to titrate 10.00 mL of 0.1450 M H2SO4 solution.

AysviL [449]

Answer:

0.1357 M

Explanation:

(a) The balanced reaction is shown below as:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

(b) Moles of $H_2SO_4$ can be calculated as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $H_2SO_4$ :

Molarity = 0.1450 M

Volume = 10.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 10×10⁻³ L

Thus, moles of $H_2SO_4$ :

$Moles=0.1450 \times {10\times 10^{-3}}\ moles$

Moles of $H_2SO_4$ = 0.00145 moles

From the reaction,

1 mole of $H_2SO_4$ react with 2 moles of NaOH

0.00145 mole of $H_2SO_4$ react with 2*0.00145 mole of NaOH

Moles of NaOH = 0.0029 moles

Volume = 21.37 mL = 21.37×10⁻³ L

Molarity = Moles / Volume = 0.0029 / 21.37×10⁻³ M = 0.1357 M

7 0

3 years ago