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Tpy6a [65]
4 years ago
15

An object is acted upon by a force of 22 newtons to the right and a force of 13 newtons to the left. What is the magnitude and d

irection of the
net force?
ОА
9 newtons to the right
ОВ.
9 newtons to the left
Ос.
22 newtons to the right
OD.
35 newtons to the right
O E.
35 newtons to the left
Chemistry
1 answer:
umka2103 [35]4 years ago
5 0
22 Newton’s to the right but I’m not to sure make sure to check other answers!
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A cubic box with sides of 20.0 cm contains 2.00 × 1023 molecules of helium with a root-mean-square speed (thermal speed) of 200
Anarel [89]

Answer:

1.133 kPa is the average pressure exerted by the molecules on the walls of the container.

Explanation:

Side of the cubic box = s = 20.0 cm

Volume of the box ,V= s^3

V=(20.0 cm)^3=8000 cm^3=8\times 10^{-3} m^3

Root mean square speed of the of helium molecule : 200m/s

The formula used for root mean square speed is:

\mu=\sqrt{\frac{3kN_AT}{M}}

where,

= root mean square speed

k = Boltzmann’s constant = 1.38\times 10^{-23}J/K

T = temperature = 370 K

M = mass helium = 3.40\times 10^{-27}kg/mole

N_A = Avogadro’s number = 6.022\times 10^{23}mol^{-1}

T=\frac{\mu _{rms}^2\times M}{3kN_A}

Moles of helium gas = n

Number of helium molecules = N =2.00\times 10^{23}

N = N_A\times n

Ideal gas equation:

PV = nRT

Substitution of values of T and n from above :

PV=\frac{N}{N_A}\times R\times \frac{\mu _{rms}^2\times M}{3kN_A}

PV=\frac{N\times R\times \mu ^2\times M}{3k\times (N_A)^2}

R=k\times N_A

PV=\frac{N\times \mu ^2\times M}{3}

P=\frac{2.00\times 10^{23}\times (200 m/s)^2\times 3.40\times 10^{-27} kg/mol}{3\times 8\times 10^{-3} m^3}

P=1133.33 Pa =1.133 kPa

(1 Pa = 0.001 kPa)

1.133 kPa is the average pressure exerted by the molecules on the walls of the container.

3 0
3 years ago
You’ll soon make two pieces of ice with the approximate shapes shown in the image. Both pieces of ice have the same mass and vol
Montano1993 [528]
Is there an image? I'll stay to help you.
3 0
4 years ago
2. 1.5 moles of AgNO3 reacts with 0.5 mole of Mg3P2. Calculate the moles of excess
kicyunya [14]

Answer:

No of Moles in excess at the end of the reaction  is 0.25 moles

Explanation:

AgNO3  +   Mg3P2  → Ag3P + Mg(NO3)2

Balancing the equation we get

6AgNO3  +   Mg3P2  → 2Ag3P + 3Mg(NO3)2

6 moles of AgNO3 needs 1 mole of Mg3P2

using unitary method

AgNO3 = \frac{1}{6}*Mg3P2

1.5 AgNO3 = \frac{1}{6}*1.5

                  = 1/4 = 0.25moles of Mg3P2

So 1.5 Moles of AgNO3 requires 0.25Mg3P2 for complete reaction but we have 0.5Moles of Mg3P2 available Therefore Mg3P2 is in excess

No of Moles in excess at the end of the reaction = 0.5 - 0.25 = 0.25moles

5 0
3 years ago
PLEASE HELP!! WILL MARK BRAINLIEST!! TIME IS RUNNING OUT!!!
Juli2301 [7.4K]

<span>We can use the heat equation,
Q = mcΔT </span>

<span>

Where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g</span>⁻¹ °C⁻<span>¹) and ΔT is the temperature difference (°C).</span>

 

According to the given data,

Q = 300 J 

m = 267 g

<span> c = ?
ΔT = 12 °C</span>

 

By applying the formula,

<span>300 J = 267 g x c x 12 °C
       c = 0.0936 J g</span>⁻¹ °C⁻<span>¹

Hence, specific heat of the given substance is </span>0.0936 J g⁻¹ °C⁻¹.

 


6 0
4 years ago
Help i just need some help with thos
Alona [7]
Okay give me a few minutes and I will send u a screen shot of my answers.
5 0
3 years ago
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