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Questions in chemistry

📝 Answered - Type the correct answer in the box. The solubility of calcium carbonate is $14 \frac{\text { milligrams }}{\text { liter }}$. This rate means that 14 milligrams of calcium carbonate can dissolve in 1 liter of water. How much water would be required to fully dissolve 11 grams of calcium carbonate? Express your answer to the correct number of significant figures. One milligram is equal to 0.001 grams. It would take $\square$ liters of water to fully dissolve 11 grams of calcium carbonate.

📝 Answered - Calculate the pH at [tex]$25^{\circ} C$[/tex] of a 0.58 M solution of pyridinlum chloride [tex]$\left( C _5 H _5 NHCl \right)$[/tex]. Note that pyridine [tex]$\left( C _5 H _5 N\right)$[/tex] is a weak base with a [tex]$pK_b$[/tex] of 8.77. Round your answer to 1 decimal place.

📝 Answered - James Chadwick’s experiment in 1932 provided evidence for the existence of _______ within the nucleus

📝 Answered - Calculate the pH at $25^{\circ} C$ of a 0.68 M solution of sodium propionate $\left( NaC _2 H _5 CO _2\right)$. Note that propionic acid $\left( HC _2 H _5 CO _2\right)$ is a weak acid with a $p K_a$ of 4.89. Round your answer to 1 decimal place. $pH=$

📝 Answered - Consider the following system at equilibrium. [tex]CaCO_3(s) \Leftrightarrow Ca^{2+}(aq)+CO_3^{2-}(aq)[/tex] The addition of which compound will cause a shift in equilibrium because of a common ion effect? A. [tex]CCl _4[/tex] B. [tex]CO _2[/tex] C. [tex]CuSO _4[/tex] D. [tex]Na _2 CO _3[/tex]

📝 Answered - What is k in the rate law equation? Rate=k[A][B] A. A pressure constant B. A volume constant C. A rate constant D. An equilibrium constant

📝 Answered - A 95.0 g sample of copper $\left( c _y=0.20 J /{ }^{\circ} C \cdot g \right)$ is heated to $82.4^{\circ} C$ and then placed in a container of water $\left(c_y=4.18 J /{ }^{\circ} C \cdot g \right)$ at $22.0^{\circ} C$. The final temperature of the water and the copper is $25.1^{\circ} C$. What was the mass of the water in the original container? Assume that all heat lost by the copper is gained by the water. Use the formulas below to help in your problem-solving. $\begin{aligned} -q_{\text {mas }} & =q_{\text {wate }} \\ -c_m m_m \Delta T_m & =c_w m_v \Delta T_w $\end{aligned}$ A. 0.246 g H _2 O B. 4.73 g H _2 O C. 84.0 g H _2 O D. 36,700 g H _2 O

📝 Answered - How many calories are released when 500 g of water cools from 95.0°C to 25.0°C? A. 4.75 x 104 cal B. 1.25 x 104 cal C. 70.0 cal D. 35.0 cal E. 3.50 x 104 cal

📝 Answered - The table below shows the freezing points of four substances. | Substance | Freezing point (°C) | | --------- | ------------------- | | benzene | 5.50 | | water | 0.00 | | butane | -138 | | nitrogen | -210 | The substances are placed in separate containers at room temperature, and each container is gradually cooled. Which of these substances will solidify before the temperature reaches [tex]$0^{\circ} C$[/tex]? A. benzene B. water C. butane D. nitrogen

📝 Answered - How does the rate law show how concentration changes affect the rate of reaction? A. The rate is expressed as the sum of the concentrations of reactants raised to some power. B. The rate is expressed as the difference between the concentrations of reactants. C. The rate is expressed as a ratio of the concentrations of reactants raised to some power. D. The rate is expressed in terms of concentrations of the reactants raised to some power.