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Sloan [31]
3 years ago
6

What can one conclude from the fact that water has a higher specific heat than sand?

Chemistry
1 answer:
miss Akunina [59]3 years ago
7 0
Sand will burn faster
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44.8% of a 250. mL acid solution is used in an experiment, what volume of acid (in mL) was used?
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Answer:

112 mL  

Explanation:

The formula for percent by volume is

\text{Percent by volume} = \dfrac{\text{Volume of solute}}{\text{Volume of solution}}\times 100 \, \%

If you have 250 mL of a solution that is 44.8 % v/v,

\begin{array}{rcl}44.8\, \% & = & \dfrac{\text{Volume of solute}}{\text{250 mL}}\times 100 \, \%\\\\44.8 \times \text{ 250 mL} & = & \text{Volume of solute} \times 100\\\text{Volume of solute} & = & \dfrac{44.8 \times 250\text{ mL}}{100}\\\\ & = & \textbf{112 mL}\\\end{array}

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3 years ago
A sample of helium gas is in a sealed rigid container. What occurs as the temperature of the sample is increased
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The electrons will move more rapidly resulting in a higher pressure even at a consistent volume
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The illustration depicts the formation of an ionic chemical bond between lithium and fluorine atoms. Why is the resulting compou
nirvana33 [79]
<span>The transferred electron from lithium to fluorine provides each atom with a full outer energy level.</span>
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During the third stage of labor what happens? (it's not C)
QveST [7]

Answer:

B

Explanation:

placenta comes after baby

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2 years ago
Consider the reaction: P(s) + 5/2 Cl2(g)PCl5(g) Write the equilibrium constant for this reaction in terms of the equilibrium con
Pani-rosa [81]

Answer: The equilibrium constant for the overall reaction is K_a\times K_b

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

a) P(s)+\frac{3}{2}Cl_2(g)\rightarrow PCl_3(g)

K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}

b) PCl_3(g)+Cl_2(g)\rightarrow PCl_5(g)

K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}

For overall reaction on adding a and b we get c

c) P(s)+\frac{5}{2}Cl_2(g)\rightarrow PCl_5(g)

K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}

K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}

The equilibrium constant for the overall reaction is K_a\times K_b

4 0
3 years ago
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