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3 years ago

At terminal velocity, which two forces are balanced? *

Chemistry

1 answer:

OlgaM077 [116]3 years ago

8 0

Answer:

friction and air resistance

at terminal velocity, the weight of the object due to gravity is balanced by the frictional forces, and the resultant force is zero.

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As an electron in an atom moves from the ground state to the excited state, the electron (1) gains energy as it moves to a highe

yawa3891 [41]

Electron requires energy to move from ground state to excited state(higher energy levels). Hence option 1 is correct. Hope this helps!

5 0

3 years ago

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

Alik [6]

Answer : The rate for trial 5 will be $4.25\times 10^{-2}Ms^{-1}$

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:

$6.0\times 10^{-4}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-3}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

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

Expression for rate law for fourth observation:

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

Dividing 1 from 2, we get:

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

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

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:

$6.0\times 10^{-4}=k(0.20)^2(0.20)^0(0.20)^1$

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

Thus, the value of the rate constant 'k' for this reaction is $7.5\times 10^{-2}M^{-2}s^{-1}$

Now we have to calculate the rate for trial 5 that starts with 0.90 M of reagent A, 0.60 M of reagents B and 0.70 M of reagent C.

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

$\text{Rate}=(7.5\times 10^{-2})\times (0.90)^2(0.60)^0(0.70)^1$

$\text{Rate}=4.25\times 10^{-2}Ms^{-1}$

Therefore, the rate for trial 5 will be $4.25\times 10^{-2}Ms^{-1}$

3 0

3 years ago

True or false applied chemistry does not have a specific goal or application

melisa1 [442]

Answer:

False. The goal of chemistry is to accumulate knowledge. Sometimes true. Biochemistry involves ... Sometimes true. Applied Chemistry is used to attain specific goals ... Applied Chemistry.

Explanation:

I really hope this helped :)

7 0

3 years ago

Validate how valence electrons determine the chemical reactivity of an element: give 2 examples of elements with high reactivity

Ostrovityanka [42]

Nitrogen is more reactive than oxygen and oxygen than chlorine

8 0

3 years ago

0.102 g of an unknown(non electrolyte) compound is dissolved in enough water to make 100. mL of a solution and has an osmotic pr

Leni [432]

Answer:

680 g/m is the molar mass for the unknown, non electrolyte, compound.

Explanation:

Let's apply the formula for osmotic pressure

π = Molarity . R . T

T = T° absolute (in K)

R = Universal constant gases

π = Pressure

Molarity = mol/L

As units of R are L.atm/mol.K, we have to convert the mmHg to atm

760 mmHg is 1 atm

28.1 mmHg is (28.1 .1)/760 = 0.0369 atm

0.0369 atm = M . 0.082 L.atm/mol.K . 293K

(0.0369 atm / 0.082 mol.K/L.atm . 293K) = M

0.0015 mol/L = Molarity

This data means the mol of solute in 1L, but we have 100mL so

Molarity . volume = mol

0.0015 mol/L . 0.1L = 1.5x10⁻⁴ mole

The molar mass will be: 0.102g / 1.5x10⁻⁴ m = 680 g/m

4 0

4 years ago