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2 years ago

An unknown or changeable quantity is called a(n)...

1 answer:

6 0

I think the answer would be dependent variable. An unknown or changeable quantity is called a dependent variable. It <span>is what you measure in the experiment and what is affected during the experiment. Hope this answers the question. Have a nice day.</span>

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What can you predict about an element from its position in the periodic table?

Igoryamba

Yeah of-course!! It's valency by group most of the chemical property like electronegativity, ionization energy etc. by the combination of groups and periods...

8 0

2 years ago

PLEASE HELP ME!!! ASAP

shtirl [24]

Answer:

Theoretical yield of the reaction = 34 g

Excess reactant is hydrogen

Limiting reactant is nitrogen

Explanation:

Given there is 100 g of nitrogen and 100 g of hydrogen

Number of moles of nitrogen = 100 ÷ 28 = 3·57

Number of moles of hydrogen = 100 ÷ 2 = 50

Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation

N2 + 3H2 → 2NH3

From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed

Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen

⇒ We require 10·71 moles of hydrogen

But we have 50 moles of hydrogen

∴ Limiting reactant is nitrogen and excess reactant is hydrogen

From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia

Molecular weight of ammonia = 17 g

∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g

5 0

3 years ago

A certain gas is present in a 13.0 L cylinder at 1.0 atm pressure. If the pressure is increased to 2.0 atm , the volume of the g

Scorpion4ik [409]

<h3>Answer:</h3>

The gas obeys the Boyle's law

<h3>Explanation:</h3>

According to Boyle's law, the volume of a fixed mass of a gas and the pressure are inversely proportional at constant absolute temperature.
That is; $P\alpha\frac{1}{V}$
Therefore, $k=PV$ , where k is a constant
At varying volume and pressures while keeping absolute temperature constant; k = P1V1 =P2V2

In this case, we are given;

Initial Volume of 13.0 L at initial pressure of 1.0 atm

New volume of 6.5 L at new pressure of 2.0 atm

But, K = PV

Therefore,

k1 = P1V1

= 1.0 atm × 13.0 L

= 13 atm.L

k2 = P2V2

= 2.0 atm × 6.5 L

= 13 atm.L

Thus, k1=k2

Thus, the gas obeys the Boyle's law

8 0

2 years ago

A student combines 34 grams of one chemical and an unknown amount of another chemical in a closed system. A chemical reaction oc

Stells [14]

Answer:

82 grams

Step by step:

116-34=82

6 0

2 years ago

Calculate the density of O2(g) at 415 K and 310 bar using the ideal gas and the van der Waals equations of state. Use a numerica

Lera25 [3.4K]

Answer:

Explanation:

From the given information:

The density of O₂ gas = $d_{ideal} = \dfrac{P\times M}{RT}$

here:

P = pressure of the O₂ gas = 310 bar

= $310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}$

= 305.97 atm

The temperature T = 415 K

The rate R = 0.0821 L.atm/mol.K

molar mass of O₂ gas = 32 g/mol

∴

$d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}$

$d_{ideal}$ = 287.37 g/L

To find the density using the Van der Waal equation

Recall that:

the Van der Waal constant for O₂ is:

a = 1.382 bar. L²/mol² &

b = 0.0319 L/mol

The initial step is to determine the volume = Vm

The Van der Waal equation can be represented as:

$P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}$

where;

R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol

Replacing our values into the above equation, we have:

$310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}$

$310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}$

$310V^3 -44.389V^2+1.382V-0.044=0$

After solving;

V = 0.1152 L

∴

$d_{Van \ der \ Waal} = \dfrac{32}{0.1152}$

$d_{Van \ der \ Waal}$ = 277.77 g/L

We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.

5 0

2 years ago