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3 years ago

What is always true of an object with a lit of mass ?

Chemistry

2 answers:

yawa3891 [41]3 years ago

4 0

A greater force is required

sertanlavr [38]3 years ago

3 0

Such an object makes a larger dent in the fabric of space-time than an object with little mass. (It has a greater gravitational attraction than less massive objects)

A greater force is required to accelerate such an object than a less massive object

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Which of the following will stay constant, no matter if the substance is in the solid, liquid, or gas state? volume shape temper

Harman [31]

Answer : The correct option is, Mass

Explanation :

As we know that there are 3 states of matter :

Solid state : It is a state in which the particles are closely packed and does not have any space between them. This state have a definite shape and volume.

Liquid state : It is a state in which the particles are present in random and irregular pattern. The particles are closely arranged but they can move from one place to another. This state have a definite volume but does not have a fixed shape.

Gaseous state : It is a state in which the particles are loosely arranged and have a lot of space between them. This state have indefinite volume as well as shape.

If we are taking 100 grams of ice then after melting its mass remains same but its shape, volume and temperature will be changed and after evaporation its mass remains same but its shape, volume and temperature will be changed.

Hence, the mass will stay constant, no matter if the substance is in the solid, liquid, or gas state.

4 0

3 years ago

Read 2 more answers

For the reaction 2N2O5(g) <---> 4NO2(g) + O2(g), the following data were colected:

KonstantinChe [14]

Answer:

a) The reaction is first order, that is, order 1. Option C is correct.

b) The half life of the reaction is 23 minutes. Option B is correct

c) The initial rate of production of NO2 for this reaction is approximately = (3.7 × 10⁻⁴) M/min. Option has been cut off.

Explanation:

First of, we try to obtain the order of the reaction from the data provided.

t (minutes) [N2O5] (mol/L)

0 1.24x10-2

10 0.92x10-2

20 0.68x10-2

30 0.50x10-2

40 0.37x10-2

50 0.28x10-2

70 0.15x10-2

Using a trial and error mode, we try to obtain the order of the reaction. But let's define some terms.

C₀ = Initial concentration of the reactant

C = concentration of the reactant at any time.

k = rate constant

t = time since the reaction started

T(1/2) = half life

We Start from the first guess of zero order.

For a zero order reaction, the general equation is

C₀ - C = kt

k = (C₀ - C)/t

If the reaction is indeed a zero order reaction, the value of k we will obtain will be the same all through the set of data provided.

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = (0.0124 - 0.0092)/10 = 0.00032 M/min

At t = 20 minutes, C = 0.0068 M

k = (0.0124 - 0.0068)/20 = 0.00028 M/min

At t = 30 minutes, C = 0.0050 M

k = (0.0124 - 0.005)/30 = 0.00024 M/min

It's evident the value of k isn't the same for the first 3 trials, hence, the reaction isn't a zero order reaction.

We try first order next, for first order reaction

In (C₀/C) = kt

k = [In (C₀/C)]/t

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = [In (0.0124/0.0092)]/10 = 0.0298 /min

At t = 20 minutes, C = 0.0068 M

k = 0.030 /min

At t = 30 minutes, C = 0.0050 M

k = 0.0303

At t = 40 minutes

k = 0.0302 /min

At t = 50 minutes,

k = 0.0298 /min

At t = 60 minutes,

k = 0.031 /min

This shows that the reaction is indeed first order because all the answers obtained hover around the same value.

The rate constant to be taken will be the average of them all.

Average k = 0.0302 /min.

b) The half life of a first order reaction is related to the rate constant through this relation

T(1/2) = (In 2)/k

T(1/2) = (In 2)/0.0302

T(1/2) = 22.95 minutes = 23 minutes.

c) The initial rate of production of the product at the start of the reaction

Rate = kC (first order)

At the start of the reaction C = C₀ = 0.0124M and k = 0.0302 /min

Rate = 0.0302 × 0.0124 = 0.000374 M/min = (3.74 × 10⁻⁴) M/min

3 0

3 years ago

You have two compounds that you have spotted on the TLC plate. One compound is more polar than the other. You ran the TLC plate

goldenfox [79]

Answer:

we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf

Explanation:

when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf

The speed of the polar spot depends largely on the level of polarity, an increase in the polarity will see both spots of Neat hexane run when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate

3 0

3 years ago

Using the phase diagram for H2O what phase is water in at 1 atm pressure and 50oC? A. It is in the solid phase B. It is at its m

galina1969 [7]

Answer:

D.It is in the gas phase

Explanation:

hope it helps .

6 0

3 years ago

Chemistry gcse

mart [117]

Yes, anything with carbonate, hydrogen carbonate (bicarbonate) at the end is a carbonate.

Examples:NaHCO3 (Sodium hydrogen carbonate or Sodium bicarbonate)Na2CO3 (Sodium carbonate)

8 0

3 years ago