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2 years ago

If a position time graph has a constant slope, what can you infer about the acceleration?

Chemistry

1 answer:

gogolik [260]2 years ago

5 0

Answer:

Acceleration is zero.

Explanation:

The slope of a position time graph gives the velocity of the body.

If the slope is constant means the velocity is constant.

Now, acceleration is the measure of the change in velocity of a body over a given time interval.

So, the acceleration of a body is directly proportional to the change in velocity of the body.

If there is no change in velocity, this means that the acceleration of the body is zero.

Here, the slope is a constant implying that the velocity is a constant. So, there is no change in velocity. This implies that the acceleration is zero for the body in the given time interval.

Thus, if a position time graph has a constant slope, one can infer that the acceleration is zero.

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I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

3 0

3 years ago

A piece of wood and a piece of steel are at the same temperature; however, the steel feels hotter. This is because the steel has

Anika [276]

B- Greater thermal conductivity

6 0

3 years ago

Read 2 more answers

Why is the answer C for this problem?

pshichka [43]

Answer:

$\boxed{\text{(C) X}$_{3}$P$_{2}}$

Explanation:

Step 1. Identify the Group that contains X

We look at the consecutive ionization energies and hunt for a big jump between them

$\begin{array}{crc}n & IE_{n} & IE_{n} - IE_{n-1}\\1 & 730 & \\2 & 1450 & 720\\3 & 7700 & 6250\\4 & 10500 & 2800\\\end{array}$

We see a big jump between n = 2 and n = 3. This indicates that X has two valence electrons.

We can easily remove two electrons, but the third electron requires much more energy. That electron must be in the stable, filled, inner core.

So, X is in Group 2 and P is in Group 15.

Step 2. Identify the Compound

X can lose two valence electrons to reach a stable octet, and P can do the same by gaining three electrons.

We must have 3 X atoms for every 2 P atoms.

The formula of the compound is $\boxed{\text{X}$_{3}$P$_{2}}$}$ .

4 0

2 years ago

Which clue can be used to identify a chemical reaction as a decomposition reaction?

inn [45]

Answer:

plzz mark brainliast:

Explanation:

clue: The reaction has a single reactant.

8 0

3 years ago

Read 2 more answers

Find the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.110 atm to rise

USPshnik [31]

Answer:

t = 37.1 s

Explanation:

The equation for the reaction is given as;

2 N2O5(g) --> 4 NO2 + O2

Initial: 0.110 - -

change: -2x +4x +x

Final: 0.110 - 2x +4x +x

But final = 0.150atm;

0.110 - 2x + 4x + x = 0.150 atm

3x = 0.150 - 0.110

x = 0.0133 atm

Pressure in reactant side;

0.110 - 2x

0.110 - 2 (0.0133) = 0.0834 atm

The integral rate law expression is given as;

ln ( [A] / [Ao] ) = -kt

k = rate constant = 7.48*10^-3*s-1

ln (0.0834/0.11) = (7.48*10^-3) t

upon solving, t = 37.1 s

3 0

2 years ago