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3 years ago

An object that is already moving can . . .

1 answer:

4 0

An object that is not already moving will begin to move in the direction of the larger force. An object that is already moving will change its speed and/or its direction.

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Complete the following table based on known properties of ionic and covalent compounds.

TiliK225 [7]

Answer:

structure of KF - ionic

Explanation:

6 0

2 years ago

What is AgNO3 + KCI-KNO3+AgCI. What type of reaction does this represent

PIT_PIT [208]

<span>C4H10 + 6.5 O2 ----> 4CO2 + 5H2O

2C4H10 + 13 O2 ----> 8CO2 + 10H2O

1. Count the C on the left (4), put a 4 where the C on the right.

2. Count the H on the left (1), you have two on the right, so you multimply this two by 5. Put the 5 in front of the H2O

3. Count the O on the right. You have 4*2 + 5 = 13. You have two on the left, so you need 6.5 on the left.

4. Now multiply everything on the equation by two so you have nice integer numbers.

5. check you have the same amount of everything on each side.
Example C: left 8, right 8, etc.
I hope this helps. :)</span><span>
</span>

6 0

3 years ago

10 m3 of carbon dioxide is originally at a temperature of 50 °C and pressure of 10 kPa. Determine the new density and volume of

Dimas [21]

Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and $7.2m^3$ respectively.

Explanation :

First we have to calculate the new or final volume of carbon dioxide gas.

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 10 kPa

$P_2$ = final pressure of gas = 15 kPa

$V_1$ = initial volume of gas = $10m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $50^oC=273+50=323K$

$T_2$ = final temperature of gas = $75^oC=273+75=348K$

Now put all the given values in the above equation, we get:

$\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}$

$V_2=7.2m^3$

The new volume of carbon dioxide gas is $7.2m^3$

Now we have to calculate the new density of carbon dioxide gas.

$PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}$

Formula for new density will be:

$\rho_2=\frac{P_2M}{RT_2}$

where,

$P_2$ = new pressure of gas = 15 kPa

$T_2$ = new temperature of gas = $75^oC=273+75=348K$

M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

$\rho$ = new density

Now put all the given values in the above equation, we get:

$\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}$

$\rho_2=0.2281g/L$

The new density of carbon dioxide gas is 0.2281 g/L

5 0

3 years ago

A particle has 5 protons, 6 neutrons, and 5 electrons. What is its net charge.

Trava [24]

It would have a zero charge because it is a neutral atom. The number of electrons which is a negative charge is equal to the number of protons which is positive, they will cancel each other out hence meaning it will become neutral

4 0

3 years ago

A gas sample has volume 3.00 dm3 at 101

gavmur [86]

Answer:

V₂ = 21.3 dm³

Explanation:

Given data:

Initial volume of gas = 3.00 dm³

Initial pressure = 101 Kpa

Final pressure = 14.2 Kpa

Final volume = ?

Solution;

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

Now we will put the values in formula,

P₁V₁ = P₂V₂

101 Kpa × 3.00 dm³ = 14.2 Kpa × V₂

V₂ = 303 Kpa. dm³/ 14.2 Kpa

V₂ = 21.3 dm³

Explanation:

3 0

3 years ago