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How does particle motion effect pressure

Alex

The pressure is a result of the motion of particles

8 0

3 years ago

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A certain orbital of the hydrogen atom has n = 4 and l = 2 what are the possible values of ml for this orbital?

Nadya [2.5K]

The ml is also called as the magnetic quantum number. The value of ml can range from –l to +l including zero. Hence all of the possible values for ml given that l = 2 are:

3 0

3 years ago

A 59.1g sample of aluminum is put into a calorimeter (see sketch at right) that contains 250.0g of water. The aluminum sample st

Rainbow [258]

Answer:

The specific heat capacity of aluminum according to this experiment is 0.863 J/g°C

Explanation:

Step 1: Data given

Mass of aluminium = 59.1 grams

Mass of water = 250.0 grams

Initial temperature of aluminium = 91.3 °C

Initial temperature of water = 16.0 °C

Final temperature = 19.5 °C

Pressure remains constant

Specific heat capacity of water = 4.186 J/g°C

Step 2: Calculate specific heat of aluminium

Heat lost = heat gained

Qlost = -Q heat

Q = m*c*ΔT

heat aluminium = - heat water

m(aluminium) * c(aluminium) * ΔT(aluminium) = -m(water) * c(water) * ΔT(water)

⇒m(aluminium) = mass of aluminium = 59.1 grams

⇒c(aluminium) = the specific heat of aluminium = TO BE DETERMINED

⇒ΔT = the change in temperature = T2 -T2 = 19.5 - 91.3 = -71.8 °C

⇒ m(water) = 250.0 grams

⇒c(water) = the specific heat of water = 4.186 J/g°C

⇒ΔT = the change in temperature = T2 -T2 = 19.5 - 16.0 = 3.5 °C

59.1 * c(aluminium) * -71.8 °C = 250.0 * 4.186 J/g°C * 3.5 °C

c(aluminium) = 0.863 J/g°C

The specific heat capacity of aluminum according to this experiment is 0.863 J/g°C

3 0

3 years ago

What mass percent solution will result from dissolving 0.126 moles of AgNO3 in 475 g. of water

Kruka [31]

Answer:

The answer to your question is: 4.5 %

Explanation:

0.126 moles of AgNO3

mass percent = ?

mass of water = 475 g

Formula

weight percent = weight of solute / weight of solution x 100

Weight of solute

MW AgNO3 = 108 + 14 + (16 x 3)

= 108 + 14 + 48

= 170 g

170 g of AgNO3 ------------------- 1 mol

x --------------------- 0.126 moles

x = (0.126 x 170) / 1 = 21.42 g of AgNO3

Weight percent = 21.42/475 x 100

= 4.5 %

7 0

3 years ago

Magnesium metal burns in air to form a mixture of magnesium oxide (MgO, M = 40.31) and magnesium nitride (Mg3N2, M = 100.95). A

dedylja [7]

Answer:

26.95 %

Explanation:

Air contains the highest percentage of oxygen and nitrogen gases. Magnesium then combines with both of the gases:

$2 Mg (s) + O_2 (g)\rightarrow 2 MgO (s)$

$3 Mg (s) + N_2 (g)\rightarrow Mg_3N_2 (s)$

Firstly, find the total number of moles of magnesium metal:

$n_{Mg} = \frac{1.000 g}{24.305 g/mol} = 0.041144 mol$

Let's say that x mol react in the first reaction and y mol react in the second reaction. This means:

$x + y = 0.041144 mol$

According to stoichiometry, we form:

$n_{MgO} = x mol, n_{Mg_3N_2} = \frac{y}{3} mol$

Multiplying moles by the molar mass of each substance will yield mass. This means we form a total of:

$m_{MgO} = 40.31x g, m_{Mg_3N_2} = \frac{y}{3} 100.95 g =$

The total mass is given, so we have our second equation to solve:

$40.31x + 33.65y = 1.584$

We have two unknowns and two equations, we may then solve:

$x + y = 0.041144$

$40.31x + 33.65y = 1.584$

Express y from the first equation:

$y = 0.041144 - x$

Substitute into the second equation:

$40.31x + 33.65(0.04144 - x) = 1.584$

$40.31x + 1.39446 - 33.65x = 1.584$

$6.66x = 0.18954$

$x = 0.028459$

$y = 0.041144 - x = 0.012685$

Moles of nitride formed:

$n_{Mg_3N_2} = \frac{y}{3} = 0.0042282 mol$

Convert this to mass:

$m_{Mg_3N_2} = 0.0042282 mol\cdot 100.95 g/mol = 0.4268 g$

Find the percentage:

$\omega_{Mg_3N_2} = \frac{0.4268 g}{1.584 g}\cdot 100\% = 26.95 \%$

7 0

3 years ago