Two thermometers are calibrated, one in degrees Celsius and the other in degrees Fahrenheit.

Answer:

Q = 12.5 kJ

Explanation:

The formula used to compute heat is:

Q = H° * n

Where:

Q: heat (J or kJ)

H°: enthalpy of reaction (kJ/mol)

n: moles

Now, as noted in the comments, the question lacks completeness, here is the part that is missing:

Given:

2A + B  A2B (1)

ΔH° = – 25.0 kJ/mol

2A2B  2AB + A2 (2)

ΔH° = 35.0 kJ/mol

Using these two reactions, we can determine the heat change.

Using the above two reactions, we need to establish the overall reaction (the one presented in the question), so let’s combine (1) and (2):

2A + B -------> A2B   H°1 = -25 kJ/mol

2A2B --------> 2AB + A2   H°2 = 35 kJ/mol

When we add these equations, one A2B cancels out with one A2B from reaction 2, thus, we have:

2A + B + 2A2B -------> 2AB + A2

So, for the enthalpy, the values are summed:

H°3 = -25 + 35 = 10 kJ/mol

Now we can calculate the heat:

Q = 10 * 2.5 = 25 kJ

However, as we have 2A in the reaction, it does not maintain a 1:1 mole ratio, instead, it is 1:2, which requires us to adjust; thus:

Q = 25 / 2 = 12.5 kJ

Answer:

The snowball's speed after the impact is 3 m/s

Explanation:

Given the following:

mass of each ball

m₁ = 0.4 Kg

m₂ = 0.6 Kg

initial speed of both balls = v₁ = 15 m/s

Speed of 1 Kg mass post-collision =?

Applying conservation of momentum

m₁ v₁ - m₂ v₁ = (m₁+m₂) V

A negative velocity indicates that the second ball moves in the opposite direction.

0.4 x 15 - 0.6 x 15 = (1) V

Therefore,

V = - 3 m/s

Consequently,

The snowball's speed following the collision is 3 m/s

Response:

0.9 cm

Clarification:

The following illustrates the calculation of the combined rod's length increase:

As established

Length increase = expansion of aluminum rod + expansion of steel rod

= 0.9 cm

We simply summed the expansions of both the aluminum and steel rods to determine the overall increase in the joined rod's length, which must be factored in

Answer:

Explanation:

Amount of gold deposited = 0.5 g

Gold's molar mass = 197 g/mol

Time duration, t = 6 hours

= 6 × 3600

= 12600 s

Calculation of moles: mass/molar mass

= 0.5/197

= 0.00254 mole

Assuming

Au --> Au+ + e-

Faraday's constant = 9.65 x 10^4 C mol-1

Charge, Q = 96500 × 0.00254

= 244.924 C

Relation: Q = I × t

Thus, I = 244.924/12600

= 0.011 A

= 11.34 mA.

Response: The spring constant is 25 N/m.

Details:

The body’s mass is 25 g, which converts to 0.025 kg (since 1 kg = 1000 g).

The total oscillations are 20 in 4 seconds.

Oscillations per second =

Spring's frequency of vibration is =

The spring constant 'k' can be derived from the relationship involving frequency, mass, and spring constant.

The spring constant is 25 N/m.