What two states of matter are pictured in the image below?

7. Which of the following is NOT always true about a synthesis reaction!

Rainbow [258]

Answer:

C

Explanation:

the formula is a + b = ab

5 0

2 years ago

Ice has a specific heat of 2.093Jg ∘C, liquid water's specific heat is 4.184Jg ∘C, and steam has a specific heat of 1.864Jg ∘C.

Ivan

Answer:

The liquid phase will have the lowest temperature change upon heating.

Explanation:

Assuming no phase change due to heating, we know that the temperature change, is proportional to the mass heated, being the proportionality constant a quantity that depends on the material, and represents the resistance of the material to change the temperature, called specific heat.

So, if we assume that the mass is the same for the three phases, and that the amount of heat supplied is also the same,the phase with the highest specific heat will have the lowest temperature change.

So, the liquid phase will be the one that exhibits this behavior, as the specific heat of liquid water (4.184 J/gºC) is the highest among the three phases.

8 0

2 years ago

Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction

Norma-Jean [14]

Answer:

The ball would hit the floor approximately $0.55\; \rm s$ after leaving the table.

The ball would travel approximately $5.5\; \rm m$ horizontally after leaving the table.

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Let $\Delta h$ denote the change to the height of the ball. Let $t$ denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let $v_0(\text{vertical})$ denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible: $\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t$ .

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was $v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}$ .

The height of the table was $1.5\; \rm m$ . Therefore, after hitting the floor, the ball would be $1.5\; \rm m \!$ below where it was before leaving the table. Hence, $\Delta h = -1.5\;\rm m$ .

The equation becomes:

$\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}$ .

Solve for $t$ :

$\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55$ .

In other words, it would take approximately $0.55\; \rm s$ for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at $v(\text{horizontal}) =10\; \rm m \cdot s^{-1}$ ) until the ball hits the floor.

The ball was in the air for approximately $t = 0.55\; \rm s$ and would have travelled approximately $v(\text{horizontal})\cdot t \approx 5.5\;\rm m$ horizontally during the flight.