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icang [17]
3 years ago
7

Convert 5.626 1 to cl.

Chemistry
1 answer:
igomit [66]3 years ago
5 0

Answer:

562.6

Explanation:

This is because in order to convert 5.626 liters to centileters, you have to multiply the volume value by 100.

Hope this helps!

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The reaction type represented by AB ---> A + B is known as
nordsb [41]

Answer:

A. synthesis

Explanation:

8 0
3 years ago
Information related to a titration experiment is given in the balanced equation and table below.
Scorpion4ik [409]

Answer:

0.24M

Explanation:

The equation for the reaction is given below:

H2SO4 + 2KOH → K2SO4 + 2H2O

From the equation above, we obtained the following information:

nA (mole of acid) = 1

nB (mole of base) = 2

Data obtained from the question include:

Va (volume of the acid) = 12mL

Ca (concentration of the acid) =?

Vb (volume of the base) = 36mL

Cb (concentration of the base) = 0.16 M

The Ca (concentration of the acid) can be obtained as follow:

CaVa/CbVb = nA/nB

Ca x 12 / 0.16 x 36 = 1 /2

Cross multiply to express in linear form as shown below:

Ca x 12 x 2 = 0.16 x 36

Divide both side by 12 x 2

Ca = 0.16 x 36/ 12 x 2

Ca = 0.24M

Therefore, the concentration of the acid is 0.24M

6 0
3 years ago
A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
Igoryamba

Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

          V_{2} = 8.34 L,    V_{1} = 2.67 L

Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

    = -2818.68 J

Hence, work for path A is -2818.68 J.

  • For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

                   = -1 atm \times (8.34 L - 2.67 L)  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = -\frac{101.33 J}{1 atm L} \times 5.67 atm L

            = -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

3 0
3 years ago
Someone please help will mark as brainliest
stich3 [128]

Answer:

A. 2Fe + 3Cl2 --> 2FeCl3

B. 2Fe + 3O2 --> 2Fe2O3

C. C4H10O + 4O2 --> 4CO2 + 5H2O

D. C7H16 + 11O2 --> 7CO2 + 8H2O

Explanation:

7 0
3 years ago
In degrees celsius, the boiling point of water is ___ degrees higher than the freezing point. (Apex)
mixer [17]

The freezing point of water in Celsius is 0 degrees and the boiling point is 100 degrees so the answer would be 100 degrees

5 0
2 years ago
Read 2 more answers
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