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2 years ago

Billy the friendly Robot uses 50 N of force to lift a box 2 meters in the air. How much work did Billy perform?

Chemistry

1 answer:

nexus9112 [7]2 years ago

4 0

Answer:

Work done, W = 100 J

Explanation:

We have, Billy the friendly Robot uses 50 N of force to lift a box 2 meters in the air.

It is required to find the work done by Billy.

Work done by an object is given in terms of force and displacement. The formula used to find the work done is given by :

$W=Fd\\\\W=50\ N\times 2\ m\\\\W=100\ J$

So, the work performed by Billy is 100 J.

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Drawing a conclusion derived from knowledge of an established principle is called _____. inductive reasoning abductive reasoning

san4es73 [151]

a.)Inductive Reasoning

is the answer

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Given that Kp [NOTE: Kp!!!!] = 1.39 at 400 ºC for the reaction, P4(g) <=> 2 P2(g), which answer best describes the reactio

Igoryamba

Answer:

The reaction will proceed to the left to attain equilibrium.

Explanation:

The question is missing but I guess it must be about <em>how the reaction will proceed to attain equilibrium.</em>

First, we have to calculate the partial pressures using the ideal gas equation.

$pP_{4}=\frac{2.50mol\times (0.08206atm.L/mol.K)\times 673K}{25.0L} =5.52atm$

$pP_{2}=\frac{1.50mol\times (0.08206atm.L/mol.K)\times 673K}{25.0L}=3.31atm$

Now, we have to calculate the reaction quotient (Qp).

$Qp=\frac{pP_{2}^{2}}{pP_{4}} =\frac{3.31^{2} }{5.52} =1.98$

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2 years ago

Question 5 of 5

grandymaker [24]

Answer:

Explanation:

the SI unit for density is the kilogram per Cubic meter

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2 years ago

sing any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0°C for the following reaction

prisoha [69]

Answer:

2.76 × 10⁻¹¹

Explanation:

I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.

1. Calculate the free energy of formation of CCl₄

C(s)+ 2Cl₂(g)→ CCl₄(g)

ΔG°/ mol·L⁻¹: 0 0 -65.3

ΔᵣG° = ΔG°f(products) - ΔG°f(reactants) = -65.3 kJ·mol⁻¹

2. Calculate K

$\text{The relationship between $\Delta G^{\circ}$ and K is}\\\Delta G^{\circ} = -RT \ln K$

T = (25.0 + 273.15) K = 298.15 K

$\begin{array}{rcl}-65 300 & = & -8.314 \times 298.15 \ln K \\65300& = & 2479 \ln K\\26.34 & = & \ln K\\K& = & e^{26.34}\\&= & \mathbf{2.76 \times 10}^{\mathbf{11}}\\\end{array}$

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2 years ago

In an organic structure, you can classify each of the carbons as follows: Primary carbon (1o) = carbon bonded to just 1 other ca

Alla [95]

The question is incomplete, complete question is :

In an organic structure, you can classify each of the carbons as follows: Primary carbon (1°) = carbon bonded to just 1 other carbon group Secondary carbon (2°) = carbon bonded to 2 other carbon groups Tertiary carbon (3°) = carbon bonded to 3 other carbon groups Quaternary carbon (4°) = carbon bonded to 4 other carbon groups How many carbons of each classification are in the structure below? How many total carbons are in the structure? How many primary carbons are in the structure? How many secondary carbons are in the structure? How many tertiary carbons are in the structure? How many quaternary carbons are in the structure?

Structure is given in an image?

Answer:

There are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.

Explanation:

Total numbers of carbon = 10

Number of primary carbons that is carbon joined to just single carbon atom = 6

Number of secondary carbons that is carbon joined to two carbon atoms = 1

Number of tertiary carbons that is carbon joined to three carbon atoms = 2

Number of quartenary carbons that is carbon joined to four carbon atoms = 1

So, there are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.

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2 years ago