Ask question

Ask question

All categories

7 days ago

15

Pure sodium metal placed in water will spark and ignite, as well as form bubbles and gas. What are the signs that this is a chem

ical reaction? Check all that apply.
Heat is produced.

Light is produced.

Heat is absorbed.

Bubbles appear.

A precipitate forms.

The color changes.

2 answers:

Softa [913]7 days ago

7 0

Answer;

Heat is generated
Light is emitted
Bubbles form

Explanation;

Chemical reactions are characterized by the interaction of two or more substances, leading to the formation of a new substance, and are typically irreversible.
Indicators of chemical reactions include gas production, release of energy in the form of light or heat, formation of precipitates, and color changes.

kicyunya [1K]7 days ago

6 0

Answer:

Heat is generated

Light is emitted

Bubbles form

You might be interested in

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

Softa [913]

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle $\theta$ to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

where $y_o$ represents the initial height from ground level, $v_{oy}$ is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

$v_y=v_{oy}-gt$

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of $v_o$ .

The peak height is attained when $v_y=0$ , which allows us to compute the time to reach that height.

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum height reached is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is equal to 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Continuing with the calculations for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Substituting known values yields

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

5 0

12 days ago

Compare the time period of two simple pendulums of length 4m and 16m at a place.

Ostrovityanka [942]

Answer:

The period of the pendulum measuring 16 m is double that of the 4 m pendulum.

Explanation:

Recall that the period (T) of a pendulum with length (L) is defined by:

$T=2\,\pi\,\sqrt{ \frac{L}{g} }$

where "g" denotes the local gravitational acceleration.

Since both pendulums are positioned at the same location, the value of "g" will be consistent for both, and when we compare the periods, we find:

$T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2$

Thus, the duration of the 16 m pendulum is two times that of the 4 m one.

5 0

3 days ago

What happens to the particles of a liquid when energy is removed from them?

Softa [913]

Response:

D: The distance among the particles diminishes

Clarification:

Removing energy reduces the activity of molecules, similar to how one slows down in cold temperatures (I believe).

3 0

10 days ago

A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 108.7 g of water at 60.2°C with 108

Ostrovityanka [942]

Answer: The calorimeter's heat capacity is $6.72J/g^oC$

Explanation:

This scenario assumes the amount of heat lost by the hot object equals the amount of heat gained by the cold object.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat capacity of water = $4.184J/g^oC$

$c_2$ = specific heat capacity of calorimeter =?

$m_1$ = mass of water = 108.7 g

$m_2$ = mass of calorimeter = 108.7 g

$T_f$ = final temperature of the mixture = $35.0^oC$

$T_1$ = initial temperature of the water = $60.2^oC$

$T_2$ = initial temperature of calorimeter = $19.3^oC$

Now substituting all provided values into the formula, we obtain

$(108.7g)\times (4.184J/g^oC)\times (35.0-60.2)^oC=-(108.7g)\times c_2\times (35.0-19.3)^oC$

$c_2=6.72J/g^oC$

Hence, the heat capacity of the calorimeter is $6.72J/g^oC$

3 0

9 days ago

A small block of mass 200 g starts at rest at A, slides to B where its speed is vB=8.0m/s,vB=8.0m/s, then slides along the horiz

serg [1198]

Answer

Data provided:

mass of the block = 200 g = 0.2 Kg

Velocity at A = 0 m/s

Velocity at B = 8 m/s

distance of slide = 10 m

height of the block = 4 m

calculation for the block's potential energy

P = m g h

P = 0.2 x 9.8 x 4

P = 7.84 J

kinetic energy calculated as

$KE = \dfrac{1}{2}mv^2$

$KE = \dfrac{1}{2}\times 0.2 \times 7.84^2$

$KE =6.14 J$

Work done = P - KE

work = 7.84 - 6.14

work = 1.7 J

b) using the formula v² = u² + 2 a s

0 = 8² - 2 x a x 10

a = 3.2 m/s²

ma - μ mg = 0

$\mu = \dfrac{a}{g}$

$\mu = \dfrac{3.2}{9.8}$

$\mu = 0.327$

7 0

8 days ago