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3 years ago

!!!help ASAP!!! What is true of dynamic equilibrium?

Chemistry

1 answer:

MArishka [77]3 years ago

7 0

A. All the changes to a system cancel out.

Explanation:

In a dynamic equilibrium, it looks as if nothing is happening for reversible reactions because the changes simply cancels out.

This will then appear like no change is happening in it. When external stress is introduced into the system, it will annul its effect in order to remain in equilibrium.

The system must be involved in reversible chemical processes no matter how small the extent of the reversibility.
The rate of forward process is equal to the rate of the backward process.
The system is closed and cannot exchange energy.
The free energy change of the system is zero.
There is no change in concentration of each species in equilibrium.
The system offers resistance to any change in factors.

learn more:

Equilibrium brainly.com/question/10838453

#learnwithBrainly

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How many grams of ammonia produced from 1000 grams of N2?

sleet_krkn [62]

Answer:

N2 + 3H2 ———> 2NH3

As we know 1000 grams ammonia is 58.82 moles so according to unitary method,

2 mole NH3 formed by 1 mole N2 hence 58.82 NH3 will be given by 29.41 moles N2.

No. Of moles = given mass/molar mass

Implies that

Mass of nitrogen required = 29.41*28 = 823.48 grams.

Explanation:

5 0

3 years ago

A chemistry student must write down in her lab notebook the concentration of a solution of sodium hydroxide. The concentration o

HACTEHA [7]

Answer:

1.099 gmL¯¹ ≈ 1.1 gmL¯¹

Explanation:

From the question given above, the following were obtained:

Mass of empty cylinder = 9.5 g

Mass Cylinder + NaOH = 31.92 g

Volume of solution = 20.4 mL

Concentration of solution =?

Next, we shall determine the mass of sodium hydroxide, NaOH. This can be obtained as as illustrated below:

Mass of empty cylinder = 9.5 g

Mass Cylinder + NaOH = 31.92 g

Mass of NaOH =?

Mass of NaOH = (Mass Cylinder + NaOH) – (Mass of empty cylinder)

Mass of NaOH = 31.92 – 9.5

Mass of NaOH = 22.42 g

Finally, we shall determine concentration of the solution as follow:

Mass of NaOH = 22.42 g

Volume of solution = 20.4 mL

Concentration of solution =?

Concentration = mass /volume

Concentration of solution = 22.42 / 20.4

Concentration of solution = 1.099 gmL¯¹ ≈ 1.1 gmL¯¹

Therefore, the concentration of the solution is 1.1 gmL¯¹

3 0

4 years ago

A 33.0−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h

Lostsunrise [7]

Answer:

THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K

Explanation:

Mass of alloy = 33 g

Initial temperature of alloy = 93°C

Mass of water = 50 g

Initail temp. of water = 22 °C

Heat capacity of calorimeter = 9.20 J/K

Final temp. = 31.10 °C

specific heat of alloy = unknown

specific heat capacity of water = 4.2 J/g K

Heat = mass * specific heat * change in temperature = m c ΔT

Heat = heat capcity * chage in temperature = Δ H * ΔT

In calorimetry;

Heat lost by the alloy = Heat gained by water + Heat of the calorimeter

mc ΔT = mcΔT + Heat capacity * ΔT

33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)

33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1

2042.7 C = 1911 + 83,72

C = 1911 + 83.72 / 2042.7

C = 1994.72 /2042.7

C =0.9765 J/g K

The specific heat of the alloy is 0.9765 J/ g K

5 0

3 years ago

Predict the products and then balance the equation

Aneli [31]

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

Here's the balanced equation for given Double displacement reaction ~

$\sf Pb(NO_3)_2 +2 \: KI = PbI_2 +2 \: KNO_3$

The products fored are : Lead Iodide ( PbI2 ) and Potassium Nitrate ( KNO3 )

4 0

2 years ago

Read 2 more answers

A diver exhales a bubble with volume of 250 mL at pressure of 2.4 atm and temperature of 15 C. How many gas particulate in this

erastova [34]

Answer:

1.5x10²² particulates

Explanation:

Assuming ideal behaviour, we can solve this problem by using the PV=nRT formula, where:

P = 2.4 atm

V = 250 mL ⇒ 250 / 1000 = 0.250 L

n = ?

R = 0.082 atm·L·mol⁻¹·K⁻¹

T = 15 °C ⇒ 15 + 273 = 288 K

We input the given data:

2.4 atm * 0.250 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 288 K

And solve for n:

n = 0.025 mol

Finally we calculate how many particulates are there in 0.025 moles, using Avogadro's number:

0.025 mol * 6.023x10²³ particulates/mol = 1.5x10²² particulates

6 0

3 years ago