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3 years ago

Suppose an object is moving through space in a straight line. What could cause the object to start moving in a circle?

2 answers:

8 0

It could pass by a large enough object that has enough gravity to pull the object into its orbit and the object would stay in orbit because it has centripetal force.

Blizzard [7]3 years ago

3 0

A secondary force that pulls the object toward a center point

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Which of the following statements are true of acids and bases?

gtnhenbr [62]

A. Acids increase the number of OH- ions in a solution. FALSE
B. Bases increase the number of OH- ions in a solution. TRUE
C. Acids increase the number of H+ ions in a solution. TRUE
D. Bases increase the number of H+ ions in a solution. FALSE

8 0

3 years ago

A 41.1 g sample of solid CO2 (dry ice) is added to a container at a temperature of 100 K with a volume of 3.4 L.A. If the contai

marta [7]

Answer:

Approximately 6.81 × 10⁵ Pa.

Assumption: carbon dioxide behaves like an ideal gas.

Explanation:

Look up the relative atomic mass of carbon and oxygen on a modern periodic table:

C: 12.011;
O: 15.999.

Calculate the molar mass of carbon dioxide $\rm CO_2$ :

$M\!\left(\mathrm{CO_2}\right) = 12.011 + 2\times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

Find the number of moles of molecules in that $41.1\;\rm g$ sample of $\rm CO_2$ :

$n = \dfrac{m}{M} = \dfrac{41.1}{44.009} \approx 0.933900\; \rm mol$ .

If carbon dioxide behaves like an ideal gas, it should satisfy the ideal gas equation when it is inside a container:

$P \cdot V = n \cdot R \cdot T$ ,

where

$P$ is the pressure inside the container.
$V$ is the volume of the container.
$n$ is the number of moles of particles (molecules, or atoms in case of noble gases) in the gas.
$R$ is the ideal gas constant.
$T$ is the absolute temperature of the gas.

Rearrange the equation to find an expression for $P$ , the pressure inside the container.

$\displaystyle P = \frac{n \cdot R \cdot T}{V}$ .

Look up the ideal gas constant in the appropriate units.

$R = 8.314 \times 10^3\; \rm L \cdot Pa \cdot K^{-1} \cdot mol^{-1}$ .

Evaluate the expression for $P$ :

$\begin{aligned} P &=\rm \frac{0.933900\; mol \times 8.314 \times 10^3 \; L \cdot Pa \cdot K^{-1} \cdot mol^{-1} \times 298\; K}{3.4\; L} \cr &\approx \rm 6.81\times 10^5\; Pa \end{aligned}$ .

Apply dimensional analysis to verify the unit of pressure.

4 0

3 years ago

To what Celsius temperature must 67.0 mL of krypton gas at 18.0°C be changed so the volume will triple? Assume the pressure and

Mice21 [21]

Answer: 600°C

Explanation:

This reaction is explained by Charles' law as the pressure is constant.

From the question, we obtained:

V1 = 67mL

T1 = 18°C = 18 +273 = 291K

V2 = 3V1 ( Vol is tripled) = 3x67 = 201mL

T2 =?

Applying the Charles' law,

V1 /T1 = V2 /T2

67/291 = 201 / T2

Cross multiply to express in linear form.

67xT2 = 291x201

Divide both side by 67, we have:

T2 = (291x201) /67

T2 = 873K

Converting to Celsius temperature, we have

T°C = K — 273

T°C = 873 — 273 = 600°C

8 0

3 years ago

PLS HELP I REALLY NEED IT IM ON THE VERGE OF TEARS

prohojiy [21]

The volume of O₂ : 21 L

<h3>Further explanation</h3>

Given

8.7 grams of C₂H₄

Required

Volume O₂

Solution

Reaction

C₂H₄ + 3 O₂ ⇒ 2 CO₂ + 2 H₂O

mol C₂H₄(MW= 28 g/mol) :

= mass : MW

= 8.7 g : 28 g/mol

= 0.311

From the equation, mol O₂ :

= 3/1 x mol C₂H₄

= 3/1 x 0.311

= 0.933

At STP, 1 mol gas=22.4 L, so for 0.933 mol :

= 0.933 x 22.4 L

= 20.899 L ≈ 21 L

6 0

3 years ago

Where are the smallest, lightest hydrocarbons captured in a distillation tower?

disa [49]

It is at the top of the tower

7 0

3 years ago

Read 2 more answers