All categories

2 years ago

7. A gas ______.

Chemistry

1 answer:

enot [183]2 years ago

7 0

Answer:

Explanation:

A gas is a state of matter that has no fixed shape and no fixed volume gases have lower density than other states of matter such as solids and liquids.

You might be interested in

If an ultraviolet photon has a wavelength of 77.8 nm calculate the energy of one mole ultraviolet photon.

DerKrebs [107]

Answer:

Explanation:

E = (hc)/(λ)

E = (6.624x10^(-27))Js x ((3×10^8)ms^(-1)) /

(77.8x10^(-9)m)

E = 2.55 x 10^(-11) J

7 0

2 years ago

calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$

(b) The normal boiling point of water (J·K−1·mol−1) is $-109Jmole^-^1K^-^1$

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole

Explanation:

Lets calculate

(a) - General equation -

$(\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p$ = $-5_m(\beta )+5_m(\alpha )$ = $-\frac{\Delta H}{T}$

$\alpha ,\beta$ → phases

ΔH → enthalpy of transition

T → temperature transition

$(\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p$ = $= -\frac{\Delta_fH}{T_f}$

= $\frac{-6.008kJ/mole}{273.15K}$ ( $\Delta_fH$ is the enthalpy of fusion of water)

= $-22Jmole^-^1k^-^1$

(b) $(\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}$

= $\frac{40.656kJ/mole}{373.15K}$ ( $\Delta_v_a_p_o_u_rH$ is the enthalpy of vaporization)

= $-109Jmole^-^1K^-^1$

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

$\mu(l,-5$ ° $C)-\mu(s,-5$ ° $C)=-Sm\DeltaT [\mu(l,0$

$\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)$

= 109J/mole

6 0

2 years ago

Thinner in the center that at the edges; virtual image smaller.

ipn [44]

<h3><u>Answer</u>;</h3>

Concave Lenses

<h3><u>Explanation</u>;</h3>

A concave lens is thin in the middle and thick at the edges, such that it seems to cave inwards. It spreads light rays apart producing an image smaller than the actual object.
<em><u>Images formed by a concave lens are virtual, upright, reduced in size and located on the same side of the lens as the object. Diverging lenses or concave lens always produce images that share these characteristics. The location of the object does not affect the characteristics of the image. </u></em>

3 0

2 years ago

Read 2 more answers

A 3.56 L balloon is compressed to 1.00 L at constant temperature. If the initial pressure in the balloon was 101.3 kPa, what is

Dmitrij [34]

Foh right on gang shawty I’ll be ighh

6 0

2 years ago

Here is a more complex redox reaction involving the dichromate ion in acidic solution: 3N O 2 − + 8H + + C r 2 O 7 2− → 3N O 3 −

Fantom [35]

Answer:

NO2- is the reducing agent.

Cr2O7_2- is the oxidizing agent.

H+ is neither

Explanation:

Reduction is the gain in electron. A chemical specie that undergoes reduction is called the oxidizing agent.

Oxidation is simply the loss in electrons. A chemical specie that undergoes oxidation is called the reducing agent.

Let us look at the species.

The first specie is the NO2-. In this specie, the oxidation number of nitrogen changed from +3 to +5 in NO3-. Thus we can see that there is more loss of electron to have caused an increase in the oxidation number positively. This shows an oxidation. Hence, NO2- is the reducing agent.

Let us look at the chromium. We can see that the oxidation number of chromium changed from +7 to +3.

Now we can see that it is a decrease and hence, it is a gain of electron and thus it is reduction. This means the first chromium specie is the oxidizing agent.

The hydrogen ion is simply placed there to balance the ions and hence it is neither the oxidizing nor the reducing agent.

4 0

3 years ago