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Ksenya-84 [330]
2 years ago
5

A gas has a pressure of 326 mm Hg at 20 degrees Celsius. What is the temperature at 52.4 k Pa?

Chemistry
2 answers:
Darya [45]2 years ago
5 0

Answer:

346 degrees because 326+20= 346

jonny [76]2 years ago
3 0

Answer:

it's is math that you can answer with your sister and your mother

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What is the mass of KOH found in a 785 mL of a 1.43 M solution of KOH
My name is Ann [436]

Molarity is defined as the ratio of number of moles to the volume of solution in litres.

The mathematical expression is given as:

Molarity = \frac{Number of moles}{volume of solution in litres}

Here, molarity is equal to 1.43 M and volume is equal to 785 mL.

Convert mL into L

As, 1 mL = 0.001 L

Thus, volume = 785\times 0.001 = 0.785 L

Rearrange the formula of molarity in terms of number of moles:

Number of moles =molarity\times volume of solution in litres

n = 1.43 M \times 0.785 L

= 1.12255 mole

Now,  Number of moles  = \frac{mass in g}{molar mass}

Molar mass of potassium hydroxide  = 56.10 g/mol

1.12255 mole  = \frac{mass in g}{56.10 g/mol}

mass in g = 1.12255 mole\times 56.10 g/mol

= 62.97 g

Hence, mass of KOH = 62.97 g



7 0
3 years ago
If the dehydration reaction of an alcohol is successful. What changes would be seen in the ir spectrum for the product compared
Bess [88]

If the dehydration reaction of an alcohol is successful. The changes would be seen in the IR spectrum for the product compared to the starting material are as,

  • The O-H and  C-O  band is disappear from stating material
  • The addition of a C-C double bond band in the product.

In dehydration reaction of alcohol (  O-H and  C-O bond ) contain , the water molecule ( H_{2}O ) is release from the reactant and C-C double bond is form which is known as alkene in the product .

The reactant and product have different structure. To determine the structure of the compound IR spectroscopy is used. In IR spectrum the peak corresponds to 3400-3600 cm is missing in the product of  dehydration reaction of an alcohol. It means  O-H band is disappear from stating material.

learn about IR SPECTRUM

brainly.com/question/15452269

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4 0
2 years ago
What amount of energy is required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drop
Gekata [30.6K]
This is a straightforward question related to the surface energy of the droplet. 

<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>

<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>

<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>

<span>The five smaller droplets need to have the same volume as the original. Therefore </span>

<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>

<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>

<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>

<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>

<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
7 0
3 years ago
How many mL can you make of a .225 M solution if you<br> have 458 mL of a .500 M solution available?
tigry1 [53]
Answer:225000 milliliters

Hope this helps.Sorry id you get this wrong
8 0
3 years ago
Balance the following redox equation and identify the element oxidized, the element reduced, the oxidizing agent, and the reduci
earnstyle [38]

The charge of Br changed from –1 to 0, therefore it is the element which is oxidized. Since it is oxidized then Br is also the reducing agent.

 

The charge of Mn changed from +4 to +2 therefore it is the element which is reduced. Since Mn is reduced, then MnO2 is the oxidizing agent.

 

The half –reactions are:

Br: 2Br --> Br2 + 2e-

Mn: MnO2 --> Mn2+

First balance oxygen by adding H2O:

MnO2 --> Mn2+ + 2H2O

Then balance hydrogen by adding H+ ions:

4H+ + MnO2 --> Mn2 + 2H2O

Then the appropriate electrons:

4e- + 4H+ + MnO2 --> Mn2 + 2H2O

 

Multiply the half-reaction of Br by 2 because the half-reaction of Mn has 4 electrons.

4Br --> 2Br2 + 4e-

 

Combine the two half reactions and cancel common factors:

4Br-  +  4H+  +  MnO2 --> 2Br2  +  Mn2  +  2H2O

6 0
2 years ago
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