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gladu [14]
3 years ago
9

PLS HELP ASAP NO LINKS PLS WILL MARK BRAINLIEST IF CORRECT

Chemistry
1 answer:
Colt1911 [192]3 years ago
5 0

Answer:

uranium is the answer base on what I have been taught so far

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Students were asked to identify the location of elements in the
Naddika [18.5K]

Answer:амьсгал боогдов гэж найдаж байна

Explanation:

4 0
3 years ago
Calculate the osmotic pressure associated with 50.0 g of an enzyme of molecular weight 98 g/mol dissolved in water to give 2600
andrew-mc [135]

Answer:

π = 4,882 atm

Explanation:

To calculate the osmotic pressure (π), the <em>Van´t Hoff equation</em> must be used, which is:

π x V = n x R x T

<em>Where: </em>

• π: Osmotic pressure, which is the difference between the levels of the solution and the pure solvent through a semipermeable membrane, which allows the passage of the solvent but not the solute

• V: Volume of the solution, in liters unit

• n: Number of moles of solute

• R: Constant of ideal gases, equal to 0.08206 L.atm / mol.K

• T: Absolute temperature, in Kelvin degrees

With the data you provide you can calculate the osmotic pressure by clearing it from the equation, we would be equal to:

π = (n x R x T) / V

However, all data must first be converted to the corresponding units in order to replace the values ​​in the equation.

<em>Solution volume ⇒ go from mL to L: </em>

1000 mL of solution ____ 1 L

2600 mL of solution _____ X = 2.6 L

Calculation: 2600 mL x 1 L / 1000 mL = 2.6 L

<em>Temperature ⇒ Go from ° C to K </em>

T (K) = t (° C) + 273.15 = 30.0 ° C + 273.15 = 303.15 K

<em>Number of moles of solute ⇒</em> <em>It can be calculated since we have the mass of the enzyme and its molecular mass: </em>

98.0 g of enzyme ____ 1 mol

50.0 g of enzyme _____ X = 0.510 moles

Calculation: 50.0 g x 1 mol / 98.0 g = 0.510 moles

Now, you can replace the values ​​in the Van´t Hoff equation and you will get the result:

 π = (n x R x T) / V

π = (0.510 mol x 0.08206 L.atm / mol.K x 303.15 K) / 2.6 L = 4.882 atm

Therefore, <em>the osmotic pressure will be 4,882 atm</em>

3 0
3 years ago
Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following
bearhunter [10]

Answer:

210.7~g~MgCO_3

Explanation:

We have to start with the <u>reaction</u>:

MgCO_3~->~MgO~+~CO_2

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the <u>number of moles</u> that we have in the 110.0 g of carbon dioxide, to this, we have to know the <u>atomic mass of each atom</u>:

C: 12 g/mol

O: 16 g/mol

Mg: 23.3 g/mol

If we take into account the number of atoms in the formula, we can calculate the <u>molar mass</u> of carbon dioxide:

(12*1)+(16*2)=44~g/mol

In other words: 1~mol~CO_2=~44~g~CO_2. With this in mind, we can calculate the moles:

110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2

Now, the <u>molar ratio</u> between carbon dioxide and magnesium carbonate is 1:1, so:

2.5~mol~CO_2=2.5~mol~MgCO_3

With the molar mass of MgCO_3 ((23.3*1)+(12*1)+(16*3)=84.3~g/mol. With this in mind, we can calculate the <u>grams of magnesium carbonate</u>:

2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3

I hope it helps!

8 0
3 years ago
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