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krek1111 [17]
3 years ago
15

The first particles to evaporate from a liquid are those

Chemistry
1 answer:
Bond [772]3 years ago
6 0
Particles below the surface of a liquid
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Is a bag of different colored jelly beans a solution or a mixture
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a mixture

Explanation:

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Solve for Va<br><br>MaVa=MbVb​
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Answer:

Va = (MbVb)/Ma

Explanation:

Divide both sides by Ma and voila!

8 0
2 years ago
What is the acceleration object velocity of +25m/s to rest in 5.0 s
Irina-Kira [14]

The average acceleration is -5.0 m·s⁻².

The formula for acceleration (<em>a</em>) is

a = \frac{v_{f}- v_{i}}{t}

v_{f} = 25 m·s⁻¹;  v_{i} = 0; t = 5.0 s

∴ a = \frac{0-25}{5.0} = -5.0 m·s⁻²

The negative sign tells you that the object is <em>slowing down</em>, i.e., it is <em>decelerating</em>.


4 0
3 years ago
Molarity to percent by mass. Convert 1.672 mol/L MgCl2(aq) solution to percent by mass of MgCl2 in the solution. The solution de
nignag [31]

Answer:

\%m/m=14\%

Explanation:

Hello!

In this case, since the molarity of magnesium chloride (molar mass = 95.211 g/mol) is 1.672 mol/L and we know the density of the solution, we can first compute the concentration in g/L as shown below:

[MgCl_2]=1.672\frac{molMgCl_2}{L}*\frac{95.211gMgCl_2}{1molMgCl_2}=159.2\frac{gMgCl_2}{L}

Next, since the density of the solution is 1.137 g/mL, we can compute the concentration in g/g as shown below:

[MgCl_2]=159.2\frac{gMgCl_2}{L}*\frac{1L}{1000mL}*\frac{1mL}{1.137g}=0.14

Which is also the by-mass fraction and in percent it turns out:

\%m/m=0.14*100\%\\\\\%m/m=14\%

Best regards!

6 0
2 years ago
A 44.0 g sample of an unknown metal at 99.0 oC was placed in a constant-pressure calorimeter of negligible heat capacity contain
tatiyna

Answer:

C_m=0.474\frac{J}{g\°C}

Explanation:

Hello.

In this case, since this is a system in which the water is heated up and the metal is cooled down in a calorimeter which is not affected by the heat lose-gain process, we can infer that the heat lost by the metal is gained be water, it means that we can write:

Q_m=-Q_w

Thus, in terms of masses, specific heats and temperatures we can write:

m_mC_m(T_{eq}-T_m)=-m_wC_w(T_{eq}-T_w)

Whereas the equilibrium temperature is the given final temperature of 28.4 °C and we can compute the specific heat of the metal as shown below:

C_m=\frac{-m_wC_w(T_{eq}-T_w)}{m_m(T_{eq}-T_m)}

Plugging the values in and since the density of water is 1.00 g/mL so the mass is 80.0g, we obtain:

C_m=\frac{-80.0g*4.184\frac{J}{g\°C} (28.4\°C-24.0\°C)}{44.0g(28.4\°C-99.0\°C)}\\\\C_m=0.474\frac{J}{g\°C}

Best regards!

6 0
2 years ago
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