All categories

3 years ago

56. When a person is not in touch with reality, they might have

1 answer:

amm18123 years ago

8 0

D: Schizophrenia. The individual that’s diagnoses with Schizophrenia tends to be disconnected with reality, leading to breakdowns in the world around them. A man sees a crowd of screaming, aggressive ghosts when it’s only a crowd of people waiting for a bus.

You might be interested in

Energy is transferred from the sun to Earth mainly by

expeople1 [14]

Answer:

electromagnetic waves

Explanation:

Radiation is the transfer of heat energy through space. Most of them are sent through electromagnetic waves.

5 0

2 years ago

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 these is true about the mantle?

yanalaym [24]

Answer:

Explanation:

it moves in slow convection currents, hope this helps!

6 0

3 years ago

Water Erosion: Cause and Effect

Cloud [144]

Answer:

Definitely add more points

Explanation:

I'm saying this because this is worth way more points

3 0

3 years ago

The molar mass is determined by measuring the freezing point depression of an aqueous solution. A freezing point of -5.20°C is r

Dima020 [189]

Answer:

The empirical formula is C2H4O3

The molecular formula is C4H8O6

The molar mass is 152 g/mol

Explanation:

The complete question is: An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Step 1: Data given

Mass % of Carbon = 31.57 %

Mass % of H = 5.30 %

Freezing point = -5.20 °C

10.56 grams of the compound dissolved in 25.0 grams of water

Kf water = 1.86 °C kg/mol

Step 2: Calculate moles of Carbon

Suppose 31.57% = 31.57 grams

moles C = mass C / Molar mass C

moles C = 31.57 grams / 12.0 g/mol = 2.63 moles

Step 3: Calculate moles of Hydrogen:

Moles H = 5.30 grams / 1.01 g/mol

moles H = 5.25 moles

Step 4: Calculate moles of Oxygen

Moles O = ( 100 - 31.57 - 5.30) / 16 g/mol

Moles O = 3.95 moles

Step 5: We divide by the smallest number of moles

C: 2.63 / 2.63 = 1 → 2

H: 5.25/2.63 = 2 → 4

O: 3.95/ 2.63 = 1.5 → 3

The empirical formula is C2H4O3

The molar mass of the empirical formula = 76 g/mol

Step 6: Calculate moles solute

Freezing point depression = 5.20 °C = m * 1.86

m = 5.20 / 1.86

m = 2.80 molal = 2.80 moles / kg

2.80 molal * 0.025 kg = 0.07 moles

Step 7: Calculate molar mass

Molar mass = mass / moles

Molar mass = 10.56 grams / 0.07 moles

Molar mass = 151 g/mol

Step 8: Calculate molecular formula

151 / 76 ≈ 2

We have to multiply the empirical formula by 2

2*(C2H4O3) = C4H8O6

The molecular formula is C4H8O6

The molar mass is 152 g/mol

6 0

4 years ago