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Sveta_85 [38]
3 years ago
14

When a pendulum swings, if it is not continuously pushed it will not stop eventually because some of its energy is changed into.

energy?
Chemistry
1 answer:
Vladimir79 [104]3 years ago
7 0
Well, there is kinetic energy when the object is in motion. But it will stop eventually because that energy is converted into thermal energy, or heat.
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101325 pascal

Explanation:

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Adding more water to instant coffee will make the coffee a darker color
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What is the mass of 22.4 L of H2 at STP?
Vanyuwa [196]

A. 1.01 is the right answer

Since

The formula is Pv= nRT

P=1 atm

V= 22.4 L

N= x

r= 0.0821

t = 273 k (bc it’s standard temperature)

So (1)(22.4)=(x)(0.0821)(273)

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3 years ago
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In an electroplating process, copper (ionic charge +2e, atomic weight 63.6 g/mol) is deposited using a current of 10.0 A. What m
salantis [7]

Answer : The mass of copper deposit is, 1.98 grams

Explanation :

First we have to calculate the charge.

Formula used : Q=I\times t

where,

Q = charge = ?

I = current = 10 A

t = time = 10 min = 600 sec      (1 min = 60 sec)

Now put all the given values in this formula, we get

Q=10A\times 600s=6000C

Now we have to calculate the number of atoms deposited.

As, 1 atom require charge to deposited = 2\times (1.6\times 10^{-19})  

Number of atoms deposited = \frac{(6000)}{2\times(1.6\times 10^{-19})}=1.875\times 10^{22} atoms

Now we have to calculate the number of moles deposited.

Number of moles deposited = \frac{(1.875\times 10^{22})}{(6.022\times 10^{23})}=0.03113 moles

Now we have to calculate the mass of copper deposited.

1 mole of Copper has mass = 63.5 g  

Mass of Copper Deposited = 63.5\times 0.03113 =1.98g

Therefore, the mass of copper deposit is, 1.98 grams

5 0
3 years ago
A sample of xenon occupies a volume of 736 mL at 2.02 atm and 1 °C. If the volume is changed to 416 mL and the temperature is ch
kozerog [31]

Answer:

\large \boxed{\text{4.63 atm}}

Explanation:

To solve this problem, we can use the Combined Gas Laws:

\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}

Data:

p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ =    1 °C

p₂ = ?;             V₂ = 416 mL; n₂ = n₁; T₂ =  82 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = (   1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The  new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}

6 0
3 years ago
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