Where are the transition metals on the periodic table?

Dana is trying to determine if an unknown sample is a pure substance or not. The sample has a cloudy white color, and it seems e

Firdavs [7]

Answer:

Dana filtered the sample and larger granules of the sample were left behind.

Explanation:

If a substance is pure, it will have a uniform composition throughout. It will not separate into particles of various sizes.

One of the characteristics of pure substances is that they are homogeneous. A mixture is definitely made up of particles of various sizes.

Since the particles was filtered and larger granules were left behind, the sample has been separated by a physical method (filtration). Only a mixture can be separated by physical methods. It is not a pure substance.

5 0

3 years ago

Which of the following questions is outside the scope of science?

Airida [17]

A. How religion and philosophy different
All the other choices (b-d) can be explained by science while a can be explain by history I think

4 0

3 years ago

Will mark the brainiest later for correct answers! Please show work.

Elodia [21]

Answer:

According to avogadro's law, 1 mole of every substance contains avogadro's number $6.023\times 10^{23}$ of particles and weighs equal to its molecular mass.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

$\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadros number}}$

a. moles in 14.08 g of $C_{12}H_{22}O_{11}$ = $\frac{14.08g}{342.3g/mol}=0.04113moles$

molecules in 14.08 g of $C_{12}H_{22}O_{11}$ = $0.04113\times 6.023\times 10^{23}=0.2477\times 10^{23}$

b. moles in 17.75 g of NaCl = $\frac{17.75g}{58.5g/mol}=0.3034moles$

molecules in 17.75 g of $NaCl$ = $0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}$

formula units 17.75 g of $NaCl$ = $0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}$

c. moles in 20.06 g of $CuSO_4.5H_2O$ = $\frac{20.06g}{249.68g/mol}=0.08034moles$

formula units in 20.06 g of $CuSO_4.5H_2O$ = $0.08034\times 6.023\times 10^{23}=0.4839\times 10^{23}$

7 0

3 years ago

A mixture of XO2 (P = 3.00 atm) and O2 (P = 1.00 atm) is placed in a container. This elementary reaction takes place at 27 °C: 2

sukhopar [10]

Answer:

a) $\triangle G^{0} = 7.31 kJ/mol$

b) $K_{-1} = 0.0594 m^{-1} s^{-1}$

Explanation:

Equation of reaction:

$2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)$

Initial pressure 3 1 0

Pressure change 2P 1P 2P

Total pressure = (3-2P) + (1-P) + (2P)

Total Pressure = 3.75 atm

(3-2P) + (1-P) + (2P) = 3.75

4 - P = 3.75

P = 4 - 3.75

P = 0.25 atm

Let us calculate the pressure of each of the components of the reaction:

Pressure of XO2 = 3 - 2P = 3 - 2(0.25)

Pressure of XO2 =2.5 atm

Pressure of O2 = 1 - P = 1 -0.25

Pressure of O2 = 0.75 atm

Pressure of XO3 = 2P = 2 * 0.25

Pressure of XO3 = 0.5 atm

From the reaction, equilibrium constant can be calculated using the formula:

$K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }$

$K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}$

Standard free energy:

$\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol$

b) value of k−1 at 27 °C, i.e. 300K

$K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}$

$K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}$

$K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2} }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}$

6 0

3 years ago

How is a mixture different from a compound?

kakasveta [241]

Answer:

I think the answer is d. Each substance in a mixture keeps most of its characteristic properties.

6 0

2 years ago

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