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den301095 [7]
3 years ago
6

Styrofoam has a density of 0.075 g/mL. What is the volume of 21.80 g of Styrofoam?

Chemistry
1 answer:
LekaFEV [45]3 years ago
4 0

The answer is 290.66 mL

Explanation:

The density of any substance including styrofoam is determined by two main factors: mass (atoms in the substance) and volume (space occupied). Moreover, the volume or mass of the substance can be calculated by using the density as these variables are related. Below I show the process:

density = mass / volume

mass = density x volume

volume = mass / density

volume = 21.80 g / 0.075 g/mL

volume = 290.66 mL

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For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin
erastovalidia [21]

Answer : The initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+B+C\rightarrow D+E

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1

k=7.5\times 10^{-3}M^{-2}s^{-1}

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1

\text{Rate}=3.4\times 10^{-3}Ms^{-1}

Therefore, the initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

8 0
3 years ago
Which of the Following Correctly Balances this Equation? _H2+_Cl2 --> _HCl
Scorpion4ik [409]
2 hydrogen and 2 chlorine on the reactant side(left of the arrow)
There is only 1 H and 1Cl on the products side so the balanced equation would be;
H2 + Cl2 —> 2 HCl
7 0
3 years ago
You notice that the water in your friend's swimming pool is cloudy and that the pool walls are discolored at the water line. A q
Flauer [41]

Answer:

6,78 mL of 12,0 wt% H₂SO₄

Explanation:

The equilibrium in water is:

H₂O (l) ⇄ H⁺ (aq) + OH⁻ (aq)

The initial concentration of [H⁺] is 10⁻⁸ M and final desired concentration is [H⁺] = 10^{-7,20}, thus,

Thus, you need to add:

[H⁺] = 10^{-7,2} -10^{-8,0} = 5,31x10⁻⁸ M

The total volume of the pool is:

9,00 m × 15,0 m ×2,50 m = 337,5 m³ ≡ 337500 L

Thus, moles of H⁺ you need to add are:

5,31x10⁻⁸ M × 337500 L = 1,792x10⁻² moles of H⁺

These moles comes from

H₂SO₄ → 2H⁺ +SO₄²⁻

Thus:

1,792x10⁻² moles of H⁺ × \frac{1H_{2}SO_4 mol}{2H^{+} mol} = 8,96x10⁻³ moles of H₂SO₄

These moles comes from:

8,96x10⁻³ moles of H₂SO₄ × \frac{98,1 g}{1 mol} × \frac{100 gSolution}{12 gH_{2}SO_4 } × \frac{1 mL}{1,080 g}  =

6,78 mL of 12,0wt% H₂SO₄

I hope it helps!

8 0
3 years ago
In the chemical reaction, calcium carbonate (CaCO3) was heated to form two new
noname [10]

Answer: 10.99

Explanation: because you take the Cao 13.9 and take CO2 which is 10.99 and it makes 24.8 . Which is CaCO3.

8 0
3 years ago
Using the fact that at 0.10 M, the absorbance was 0.357; and at 0.20 M, the absorbance was 0.714, the absorptivity (slope) is
olga_2 [115]

Answer:

The answer is "3.57 and 0.07".

Explanation:

Using the slop formula:

=\frac{n_2-n_1}{n_2-n_1}\\\\ =\frac{0.714-0.357}{0.20-0.10} \\\\ = ABSORPTIVITY, \ \ a=3.57 \\\\A= a\times b \times c

Given:

A=0.250

length path b=1  

from calibration it is found that

a=3.57\\\\consentation \ c=\frac{A}{a\times b} \\\\C=\frac{0.250}{3.57\times 1}

   =0.07 \ M

3 0
3 years ago
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