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irakobra [83]
3 years ago
10

In Cellulose, C6H7O2(OH)3

Chemistry
1 answer:
GarryVolchara [31]3 years ago
3 0

GB do GB fo fi ch by GB hi hi every hdhs fi saw

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What are the formulas for the ionic compounds named sodium iodide and magnesium iodide
bixtya [17]

Sodium Iodide: NaI

Magnesium Iodide: MgI2

8 0
3 years ago
Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres
yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

K_f = molal freezing point elevation constant = 1.86°C/m

m_{solute} = Given mass of solute (KCl) = 5.0 g

M_{solute} = Molar mass of solute (KCl) = 74.55 g/mol

W_{solvent} = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC

Hence, the freezing point of solution is -0.454°C

3 0
3 years ago
Is gravity a matter??
algol13

Answer:

No, gravity isn't matter

Explanation:

Gravity is a <u>force</u> that attracts matter towards the center of a physical body with mass.

3 0
3 years ago
7.02 x 10^23 molecules of X2, a ssubstance consisting of diatomic molecules, has a mass of 296 grams. Determine the atomic weigh
Fiesta28 [93]

Answer:

d. 127 g/mol.

Explanation:

Hello!

In this case, since we have the amount of molecules of this this compound, we are able to compute the moles out there by using the Avogadro's number:

mol=7.02x10^{23}molec*\frac{1mol}{6.022x10^{23}molec}=1.17mol

Which correspond to the moles of X2. Then, by using the mass we are able to compute the molar mass of X2:

MM=\frac{296g}{1.17mol}\\\\MM=254g/mol

It means that the atomic mass of X halves the molar mass of X2, which is then d. 127 g/mol.

Best regards!

4 0
2 years ago
What is the density of an object with a mass of 16 g and 3.0 ml
fomenos

Answer:

D = 5.3 g/mL

Explanation:

Density = Mass over Volume

D = m/V

Step 1: Define

D = unknown

m = 16 g

v = 3.0 mL

Step 2: Substitute and Evaluate

D = 16 g / 3.0 mL

D = 5.333333333 g/mL

Step 3: Simplify

We have 2 sig figs.

5.333333333 g/mL ≈ 5.3 g/mL

6 0
3 years ago
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