1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Stells [14]
2 years ago
5

D. Convert 5.05 x 10-5 ML to hL.

Chemistry
1 answer:
Hitman42 [59]2 years ago
4 0

Answer:

d.5.05*10^-10hl

e.0.45dg

f.7.5*10^5nm

Explanation:

d.1ml=10^-5hl

e.1dg=100000ug

f.1m=1000000000nm

You might be interested in
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
3 years ago
A galvanic cell generates a cell potential of 0.32V when operated under standard conditions according to the reaction above. Whi
Ugo [173]

The complete question is shown in the image attached to this answer.

Answer:

C

Explanation:

Let us quickly remember that the EMF of a cell under non standard conditions in given by the Nernst equation.

This equation states that;

E = E°cell - 0.592/n log Q

Where

E = EMF under non standard conditions

E°cell= standard EMF of the cell

n = number of electrons transferred

Q = reaction quotient

If the reaction quotient is greater than 1 then cell potential is less than the standard cell potential.

The cell that generates the lowest cell potential is the cell depicted in option C because Q has the greatest positive value(Q<1).

6 0
2 years ago
When one carbon atom is bonded with several other carbon atoms in a single line, then how is the structure described?
Solnce55 [7]
Its described as a Straight Branch, hope this helps :)
3 0
3 years ago
Determine whether the stopcock should be completely open, partially open, or completely closed for each activity involved with t
densk [106]

Answer:

Close to the calculated endpoint of a titration - <u>Partially open</u>

At the beginning of a titration - <u>Completely open</u>

Filling the buret with titrant - <u>Completely closed</u>

Conditioning the buret with the titrant - <u>Completely closed</u>

Explanation:

'Titration' is depicted as the process under which the concentration of some substances in a solution is determined by adding measured amounts of some other substance until a rection is displayed to be complete.

As per the question, the stopcock would remain completely open when the process of titration starts. After the buret is successfully placed, the titrant is carefully put through the buret in the stopcock which is entirely closed. Thereafter, when the titrant and the buret are conditioned, the stopcock must remain closed for correct results. Then, when the process is near the estimated end-point and the solution begins to turn its color, the stopcock would be slightly open before the reading of the endpoint for adding the drops of titrant for final observation.

3 0
3 years ago
Help me please anyone
Evgesh-ka [11]
Just do it one at a time. How is heart rate affected by exersice? Well, think about it; When you run, your heart rate goes up. When you rest, it goes down. 
Take you time, think carefully and you'll get it.
How are bean seeds affected by water with detergent in it? Well it's not pure water. it's like feeding it soap and expecting it to grow..
6 0
3 years ago
Other questions:
  • The conversion of co2 and h2o into organic compounds using energy from light is called
    13·1 answer
  • The mass of the quartz crystal shown below is 10 grams, and its volume is 4 cm3. What is the density of quartz?
    13·1 answer
  • NEED HELP ASAP
    13·1 answer
  • which of the following ions is formed when the positive hydrogen ions from an acid bond to water molecules? H- OH- H2O+ H3O+
    12·2 answers
  • Which step is not included in carbon cycle
    14·1 answer
  • In the drawing above, the hammer is acting as what type of simple machine? A. pulley B. screw C. wheel and axle D. lever
    15·1 answer
  • At constant pressure, how are the temperature and volume of a gas related?
    14·1 answer
  • How many mL (to the nearest mL) of 0.140-M KF solution should be added to 400. mL of 0.212-M HF to prepare a pH
    14·1 answer
  • What happens when a star dies
    13·2 answers
  • Refer to the table . Which plants would most likely be found in the rain forest
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!