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3 years ago

10

Which of these formulas represents 1 atom of sodium (Na) chemically combined with another element?

2 answers:

KengaRu [80]3 years ago

4 0

Answer:

B. NaCl

Explanation:

<em>Which of these formulas represents 1 atom of sodium (Na) chemically combined with another element? </em>

<em>A. PbS.</em> NO. There are not atoms of Na in this formula.

<em>B. NaCl</em>. YES. 1 atom of Na is combined with 1 atom of Cl.

<em>C. 2NaOH.</em> NO. 1 atom of Na is combined with 1 atom of O and 1 atom of H, thus, it is combined with more than 1 element.

<em>D. Na₂H₃IO₆.</em> NO. There are 2 atoms of Na in this formula.

Andru [333]3 years ago

3 0

The answer is B. because NaCl are just two elements with one atom from each element

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Which of the following describes an element?

bixtya [17]

Answer:

A substance made of two or more types of atoms

Explanation:

8 0

2 years ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

3 years ago

BRAINLIEST AND MAY DOUBLE POINTS!!!

raketka [301]

Given that:

Energy of bulb (Work ) = 30 J,

Time (t) = 3 sec

The power consumption = ?

We know that, Power can be defined as rate of doing work

Power (P) = Work(Energy supplied) ÷ time

= 30 ÷ 3

= 10 Watts

<em> The power consumption is 10 W.</em>

3 0

3 years ago

Imagine that you have a 6.50 l gas tank and a 4.50 l gas tank. you need to fill one tank with oxygen and the other with acetylen

Margarita [4]

V₁(O2) = 6.50<span> L
</span>p₁(O2) = 155 atm
V₂(acetylene) = <span>4.50 L
</span>p₂(acetylene) =?

According to Boyle–Mariotte law (At constant temperature and unchanged amount of gas, the product of pressure and volume is constant) we can compare two gases that have ideal behavior and the law can be usefully expressed as:

V₁/p₁ = V₂/p₂

6.5/155 = 4.5/p₂

0.042 x p₂ = 4.5

p₂ = 107.3 atm

7 0

3 years ago

A circular dartboard has a radius of 2 meters and a red circle in the center. Assume you hit the target at a random point. For w

Sati [7]

Answer:

1.549 m

Explanation:

Given:

The radius of the circular board, r = 2 m

The probability of hitting the red is given as 0.6

Now, this probability of hitting the red can be conclude as

0.6 = (Area of red)/ (Total area of the board)

Total area of the board = πr² = π × 2²

let the radius of the red area be R

thus, area of red circle, = πR²

on substituting the value of the area, we have

0.6 = (πR²)/ (π × 2²)

or

R² = 2.4

or

R = 1.549 m

Thus, the radius of the red circle is 1.549 m

3 0

3 years ago