Can you add more points to question pls

How is a compound different from mixtures

borishaifa [10]

Answer: a compound is a substance that is made up of more than one type of atoms bonded together, mixtures are a mix of multiple elements or compounds which haven't bonded together.

Explanation:

8 0

3 years ago

A package within package can be best defined as which of the following

qwelly [4]

Answer:

double-protected

Explanation:

If a package is inside a package, the item inside the second package is double-protected, because the outer package will keep it safe.

Hope this helps!

6 0

3 years ago

A tank has a volume of 0.1 m3 and is filled with He gas at a pressure of 5 x 106 Pa. A second tank has a volume of 0.15 m' and i

GarryVolchara [31]

Answer:

$P=5.6*10^{6} Pa$

Explanation:

Consider that, as the system is adiabatic, $U_{1}= U_{2}$ where U1 and U2 are the internal energies before the process and after that respectively.

Consider that: $U=H-PV$ , and that the internal energy of the first state is the sum of the internal energy of each tank.

So, $H_{1}-P_{1}V_{1}=H_{2}-P_{2}V_{2}\\H_{1}^{A} -P_{1}^{A}V_{1}^{A}+H_{1}^{B} -P_{1}^{B}V_{1}^{B}=H_{2}-P_{2}V_{2}$

Where A y B are the tanks. The enthalpy for an ideal gas is only function of the temperature, as the internal energy is too; so it is possible to assume: $H_{1}=H_{2}\\H_{1}^{A}+H_{1}^{B} =H_{2}$

So, $P_{1}^{A}V_{1}^{A}+P_{1}^{B}V_{1}^{B}=P_{2}V_{2}$

Isolating $P_{2}$ ,

$P_{2}=\frac{P_{1}^{A}V_{1}^{A}+P_{1}^{B}V_{1}^{B}}{V_{2}}$

$V_{2}=V_{1}^{A}+V_{1}^{B}=0.25m^{3}$

So,

$P_{2}=\frac{5000000*0.1+6000000*0.15}{0.25}=5600000Pa=5.6*10^{6} Pa$

5 0

3 years ago

The molal freezing point constant for copper is 23 °C/m. If pure copper melts at 1083°C, what will be the melting point of a bra

nikitadnepr [17]

Answer: The freezing point of brass is 1028.57°C

Explanation:

We are given:

13.4 mass percent of zinc in brass

This means that 13.4 grams of zinc is present in 100 g of brass

Mass of copper = [100 - 13.4] g = 86.6 g

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 copper = 1083°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 23°C/m

$m_{solute}$ = Given mass of solute (zinc) = 13.4 g

$M_{solute}$ = Molar mass of solute (zinc) = 65.38 g/mol

$W_{solvent}$ = Mass of solvent (copper) = 86.6 g

Putting values in above equation, we get:

$1083-\text{Freezing point of solution}=1\times 23^oC/m\times \frac{13.4\times 1000}{65.38g/mol\times 86.6}\\\\\text{Freezing point of solution}=1028.57^oC$

Hence, the freezing point of brass is 1028.57°C

8 0

3 years ago

The central atom in a certain molecule has 1 nonbonded electron pairs and 2 bonded electron pairs in its valence shell. the mole

leonid [27]

If a central atom has 1 nonbonded electron pair, and 2 bonded electron pairs, then the molecular geometry of this molecule would be bent at an angle of less than 120 degrees. In this case specifically, the electron pair geometry would be trigonal planar.

7 0

4 years ago