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SCORPION-xisa [38]
3 years ago
5

A sample of a gas at 25°C has a volume of 150 mL when its pressure is 0.947 atm. What will the temperature of the gas be at a pr

essure of 0.987 atm and changes to 144mL?
*please help*
Chemistry
1 answer:
dezoksy [38]3 years ago
6 0

Answer:

25°C

Explanation:

Combined Gas Law (P₁V₁)/T₁ = (P₂V₂)/T₂

(0.947 atm)(150 mL)/25°C = (0.987 atm)(144mL)/T₂

5.682 = 142.128/T₂

T₂ = 142.128/5.682

T₂ = 25.0137272756°C = 25°C

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How does marketing affect the consumer mindset?
RoseWind [281]

Answer:

Marketing affects the consumer mindset by leaving them with the decision of either purchasing a product or not and why it should be purchased or not.

6 0
2 years ago
What are the correct coefficients when this chemical equation is<br> balanced? *<br> P4 + 02 P2O5
finlep [7]

Answer:

What are the correct coefficients when this chemical equation is

balanced? *

P4 + 02 P2O5

<h2>1, 5, 2</h2>

Explanation:

For this reaction we have a combination reaction. Balancing Strategies: This combination reaction is a lot easier to balance and if you can get an even number of oxygen atoms on the reactants side of the equation.

3 0
2 years ago
A 60.0 g block of iron that has an initial temperature of 250. °C and 60.0 g bloc of gold that has an initial temperature of 45.
Maslowich

Answer:

The final temperature at the equilibrium is 204.6 °C

Explanation:

Step 1: Data given

Mass of iron = 60.0 grams

Initial temperature = 250 °C

Mass of gold = 60.0 grams

Initial temperature of gold = 45.0 °C

The specific heat capacity of iron = 0.449 J/g•°C

The specific heat capacity of gold = 0.128 J/g•°C.

Step 2: Calculate the final temperature at the equilibrium

Heat lost = Heat gained

Qlost = -Qgained

Qiron = -Qgold

Q=m*c*ΔT

m(iron) * c(iron) *ΔT(iron) = -m(gold) * c(gold) *ΔT(gold)

⇒with m(iron) = the mass of iron = 60.0 grams

⇒with c(iron) = the specific heat of iron = 0.449 J/g°C

⇒with ΔT(iron)= the change of temperature of iron = T2 - T1 = T2 - 250.0°C

⇒with m(gold) = the mass of gold= 60.0 grams

⇒with c(gold) = the specific heat of gold = 0.128 J/g°C

⇒with ΔT(gold) = the change of temperature of gold = T2 - 45.0 °C

60.0 *0.449 * (T2 - 250.0) = -60.0 * 0.128 * (T2 - 45.0 )

26.94 * (T2 - 250.0) = -7.68 * (T2 - 45.0)

26.94T2 - 6735 = -7.68T2 + 345.6

34.62T2 = 7080.6

T2 = 204.5 °C

The final temperature at the equilibrium is 204.6 °C

5 0
3 years ago
The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (
Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^{x}[B]^{y}[C]^{z}

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}

\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}

3 = 3^{x} \\\\x =1

Order w.r.t B : Use trials 2 and 5

\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}

\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}

4 = 2^{y} \\\\y =2

Order w.r.t C : Use trials 1 and 2

\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}

1 = (0.5)^{z}

z = 1

Therefore, order w.r.t C = 1

8 0
3 years ago
If the half life of a radioactive isotope is 5000 years, how much of the radioactive isotope in a specimen will be left after 10
marissa [1.9K]
Half life is the time taken by a radioactive isotope to decay by half its original mass. In this case, the halflife of the radioactive isotope is 5000 years. 
Initially the mass is 100 %; thus the mass that will be left will be given by;
New mass = Original mass × (1/2)^n where n is the number of half lives; 
  n = 10000/5000 = 2
New mass = 100% ×(1/2)^2
                 = 100 % × 1/4 
                 = 25%
Therefore; the mass left after 10000 years is 25% or 1/4 of the original mass.
5 0
3 years ago
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