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weeeeeb [17]
3 years ago
11

A sample of an ideal gas at 1.00 atm and a volume of 1.02 l was placed in a weighted balloon and dropped into the ocean. as the

sample descended, the water pressure compressed the balloon and reduced its volume. when the pressure had increased to 65.0 atm, what was the volume of the sample? assume that the temperature was held constant.
Chemistry
1 answer:
melisa1 [442]3 years ago
8 0
<span>0.0157 L. For this equation use the combined gas law, which states P1*V1/T1 = P2*V2/T2, but in this case we can remove the T, because T1 = T2, because the temperature is constant. Then we solve for V2 because it is the new volume. So P1*V1/P2 =V2 Plug in the variables, P1 = 100 atm, V1 = 1.02 L, P2 = 65.0 atm, and V2 = 0.0157 atm.</span>
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What is the mass in grams of 9.45*10^24 molecules of methanol (CH3OH)?
Angelina_Jolie [31]
Number of moles:

1 mole ---------- 6.02x10²³ molecules
? moles --------- 9.45x10²⁴ molecules

1 x ( 9.45x10²⁴) / 6.02x10²³ =

9.45x10²⁴ / 6.02x10²³ => 15.69 moles of CH3OH

Therefore:

Molar mass CH3OH = 32.04 g/mol

1 mole ------------ 32.04 g
15.69 moles -----  mass methanol

Mass methanol  = 15.69 x 32.04 / 1 => 502.7076 g


6 0
3 years ago
How many grams of oxygen are needed to react with 4.6 grams of titanium(IV) chloride
velikii [3]
The answer is: 0.78g
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3 years ago
150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac
Murljashka [212]

Answer:

0.175\; \rm mol \cdot L^{-1}.

Explanation:

Magnesium chloride and silver nitrate reacts at a 2:1 ratio:

\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s).

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to  \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}.

The precipitate silver chloride \rm AgCl is insoluble in water and barely ionizes. Hence, \rm AgCl\! isn't rewritten as ions.

Net ionic equation:

\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}.

Calculate the initial quantity of nitrate ions in the mixture.

\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}.

Since nitrate ions \rm {NO_3}^{-} do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol.

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be (1/2) of the concentration in the original solution.

\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}.

6 0
3 years ago
Is there an optimum amount of baking soda and vinegar to put in a plastic soda bottle rocket? How would you figure that out?
Snezhnost [94]
Add 1 tsp. of vinegar to the canister at a time, filling it almost to the top. You need to add as much vinegar to the canister as possible without the vinegar and the baking soda coming into contact when you later snap the lid onto the canister. Depending on the exact canister, this may be around 5 tsp.
3 0
3 years ago
Read 2 more answers
Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica
FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = \frac{1}{2}\times 0.52=0.26mol of lead (II) iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0
3 years ago
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