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aleksley [76]
3 years ago
13

if a 1.85g mass of zinc produces 475 mL og gas and your balloon weighs 0.574 g and the room temperature is 20.5°C. calculate the

amount of zinc needed to produce enough gas to get your balloon airborne by adding 1 mL to the required balloon volume so that its density is less than that of the surrounding air.
Chemistry
1 answer:
aleksley [76]3 years ago
7 0
<span>You will use an ideal gas equation here. Let us denote 'x' as the unknown sample of zinc if another 1 ml of it is added to make a total of 476mL gas. The ideal gas equation is PV = nRT. Assume atmospheric conditions and then substitute everything to the equation.

(1atm)(0.476L) = (1.85 + x)(0.08206 L-atm/mol-K)(20.5+273K)
x = 1.83 g of zinc</span>
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SOMEONE PLS HELP
Gekata [30.6K]

Answer:

<h2>464.85 mL</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we're finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

100.7 kPa = 100,700 Pa

95.1 kPa = 95,100 Pa

We have

V_2 =  \frac{100700 \times 439}{95100}  =  \frac{44207300}{95100}  \\  = 464.8506...

We have the final answer as

<h3>464.85 mL</h3>

Hope this helps you

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3 years ago
Which of the following statements describes sound energy?
Galina-37 [17]

Answer:

C.

Explanation:

8 0
2 years ago
How does the structure of covalent bonds affects their structure.
Black_prince [1.1K]
Covalent bond is a type of chemical bond which is formed as a result of sharing of electron pairs among the elements that are involved. The structure of the covalent bond is affected by the electronegativity of the elements involved. The molecules joined by covalent bond range in size from very small to very large polymers. There are different types of structures for covalent substances, these include: macromolecular substances, molecular substances and giant covalent structures. Strong bonds hold individual molecules together but there are negligible forces of attraction among them.
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3 years ago
Fe2O3 + CO → Fe + CO2 If 2 moles of Fe2O3 react with 9 moles of CO, how many moles of each product are formed?
Y_Kistochka [10]

Answer: B. 4 moles Fe and 3 moles CO2

Explanation:

write a balanced chemical reaction

that is FeO3 + 3 CO → 2 Fe + 3CO2

2 moles of Fe2O3 reacted with 3 x2=6 moles of Co to form Fe and CO2 therefore CO was in excess and Fe2O3 was limiting reagent.

use the mole ratio to determine the moles of each product.

that is the mole ratio 0f Fe2O3 : Fe is 1:2 therefore the moles of Fe = 2x2=4 moles

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8 0
2 years ago
Read 2 more answers
25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca
Kamila [148]

Answer:

Concentration of the original \rm KOH solution: approximately 1.05\; \rm mol \cdot dm^{-3}.

Explanation:

Notice that the concentration of the \rm HCl solution is in the unit \rm mol\cdot dm^{-3}. However, the unit of the two volumes is \rm cm^{3}. Convert the unit of the two volumes to \rm dm^{3} to match the unit of concentration.

\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}.

\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}.

Calculate the number of moles of \rm HCl molecules in that 0.0350\; \rm dm^{3} of 0.75\; \rm mol \cdot dm^{3} \rm HCl\! solution:

\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}.

\rm HCl is a monoprotic acid. In other words, each \rm HCl\! would release up to one proton \rm H^{+}.

On the other hand, \rm KOH is a monoprotic base. Each \rm KOH\! formula unit would react with up to one \rm H^{+}.

Hence, \rm HCl molecules and \rm KOH\! formula units would react at a one-to-one ratio.

{\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l).

Therefore, that 0.02625\; \rm mol of \rm HCl molecules would neutralize exactly the same number of \rm NaOH formula units. That is: n(\mathrm{NaOH}) = 0.02625\; \rm mol.

Calculate the concentration of a \rm NaOH solution where V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3} and n(\mathrm{NaOH}) = 0.02625\; \rm mol:

\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}.

7 0
2 years ago
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