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3 years ago

Describe how the properties of the compound carbon monoxide (CO) differ from the properties of the elements in the compound.

Physics

2 answers:

abruzzese [7]3 years ago

8 0

Answer:

A compound is a pure substance made of two or more elements chemically combined in a set ratio. ... If a different ratio of carbon atoms and oxygen atoms are seen in a formula, you have a different compound. For example, carbon monoxide—a gas produced in car engines—has the formula CO.

Harlamova29_29 [7]3 years ago

7 0

Answer:

Carbon monoxide is a highly poisonous, odorless, colorless, and tasteless gas. It is very flammable in air over a wide range of concentrations and burns in air with a bright blue flame . It becomes a liquid at 81.62 K.

the elements in the compound carbon and oxygen.

oxygen is also colorless, odorless and tasteless gas but it's not poisonous.

as we use it for breathing purpose whereas too much oxygen can be toxic.

carbon is solid and very brittle. diamond and graphite are two most common forms of carbon.

When elements are chemically combined, they form compounds having properties that are different from those of the uncombined elements

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2 years ago

P6: An object of mass m sits on a spring of constant k in an elevator that is accelerating upwards with acceleration a. a) In te

tankabanditka [31]

Answer:

(a). The spring compressed is $\dfrac{ma+mg}{k}$ .

(b). The acceleration is 1.5 g.

Explanation:

Given that,

Acceleration = a

mass = m

spring constant = k

(a). We need to calculate the spring compressed

Using balance equation

$kx-mg=ma$

$x=\dfrac{ma+mg}{k}$ ....(I)

The spring compressed is $\dfrac{ma+mg}{k}$ .

(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,

The compression is given by

$x=2.5\times x_{0}$

Here, acceleration is zero

So, $x=2.5\times\dfrac{mg}{k}$

We need to calculate the acceleration

Put the value of x in equation (I)

$2.5\times \dfrac{mg}{k}=\dfrac{ma+mg}{k}$

$2.5\times\dfrac{mg}{k}=\dfrac{m}{k}(a+g)$

$a=2.5g-g$

$a=1.5g$

Hence, (a). The spring compressed is $\dfrac{ma+mg}{k}$ .

(b). The acceleration is 1.5 g.

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3 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

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3 years ago