What is a chemical bond?

4. Balance the following equation:<br> _NO + __02,_NO2

Ipatiy [6.2K]

Answer:

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5 0

4 years ago

A box has a volume of 45m3 and is filled with air held at 25∘C and 3.65atm. What will be the pressure (in atmospheres) if the sa

Marina CMI [18]

Answer:

Given:

Initial pressure: $3.65\; \rm atm$ .
Volume was reduced from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ .
Temperature was raised from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ .

New pressure: approximately $3.4\times 10\; \rm atm$ ( $34\; \rm atm$ .) (Assuming that the gas is an ideal gas.)

Explanation:

Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:

Reduce the volume of the gas from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ . Calculate the new pressure, $P_1$ .
Raise the temperature of the gas from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ . Calculate the final pressure, $P_2$ .

By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)

For this gas, $V_0 = 45\; \rm m^{3}$ while $V_1 = 5.0\; \rm m^{3}$ .

Let $P_0$ denote the pressure of this gas before the volume change ( $P_0 = 3.65\; \rm atm$ .) Let $P_1$ denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between $P_1\!$ and $P_0\!$ :

$\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0$ .

In other words, because the final volume is $(1/9)$ of the initial volume, the final pressure is $9$ times the initial pressure. Therefore:

$\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm$ .

On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)

Convert the unit of the temperature of this gas to degrees Kelvins:

$T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K$ .

$T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K$ .

Let $P_1$ denote the pressure of this gas before this temperature change ( $P_1 = 32.85\; \rm atm$ .) Let $P_2$ denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at $V_2 = V_1 = 5.0\; \rm m^{3}$ .

Apply Amonton's Law to find the ratio between $P_2$ and $P_1$ :

$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}$ .

Calculate $P_2$ , the final pressure of this gas:

$\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}$ .

In other words, the pressure of this gas after the volume and the temperature changes would be approximately $3.4\times 10\; \rm atm$ .

8 0

3 years ago

What is the component concentration ratio, [no2−]/[hno2], of a buffer that has a ph of 3.90? (ka of hno2 = 7.1 × 10−4)?

sasho [114]

According to Henderson–Hasselbalch Equation,

pH = pKa + log [Nitrate] / [Nitric Acid]

As, Ka of Nitric Acid = 7.1 × 10⁻⁴
So,
pKa = -log [ 7.1 × 10⁻⁴ ]

pKa = 3.148

So, pH = 3.148 + log [Nitrate] / [Nitric Acid]

3.90 = 3.148 + log [Nitrate] / [Nitric Acid]

3.90 - 3.148 = log [Nitrate] / [Nitric Acid]

0.752 = log [Nitrate] / [Nitric Acid]

Taking Antilog on both sides,

[Nitrate] / [Nitric Acid] = 5.64

8 0

3 years ago

An object with mass m is attached to the end of a string and is raised vertically at a constant acceleration of g 6. If it has b

sattari [20]

11/10 MgL work has been done by the tension in the string. If the object has been raised a distance l from rest.

<h3>Define work done?</h3>

The definition of work is expanded so that it encompasses both the forces applied to the body and the overall displacement of the body. A constant force F is present before this block. The objective of this force is to move the body d meters in a straight line in the force's direction.

Given,

Mass of object, M

Acceleration of object, a = g/10

Distance covered vertically, L

Tension of string, T

Work done by tension in string, W

W = T × L

By applying equilibrium force on string,

T - Mg = Ma

g = gravitational acceleration

T - Mg = M (g/10)

T = M(g/10) + Mg

T = 11/10 Mg

By substituting the value in, W = T × L

W = 11/10 Mg × L

W = 11/10 MgL

∴ Work done by tension is 11/10 MgL.

To know more about work done, visit:

brainly.com/question/13662169

#SPJ4

8 0

1 year ago

The Nernst equation at 20oC is:

saw5 [17]

Answer:

a. -58 millivolts

Explanation:

The given Nernst equation is:

$E_{ion} = 58 millivolts /z \Big[ log_{10} \Big( \dfrac{[ion]_{out}}{[ion]_{in}}\Big) \Big]}$

The equilibrium potential given by the Nernst equation can be determined by using the formula:

$E_{Cl^-} = \dfrac{2.303*R*T}{ZF} \times log \dfrac{[Cl^-]_{out}} {[Cl^-]_{in}}$

where:

gas constant(R) = 8.314 J/K/mol

Temperature (T) = (20+273)K

= 298K

Faraday constant F = 96485 C/mol

Number of electron on Cl = -1

$E_{Cl^-} = \dfrac{2.303*8.314*298} {(-1)*(96845)} \times log \dfrac{100} {10}$

$E_{Cl^-} = - 0.05814 \ volts$

$\mathsf{E_{Cl^-} = - 0.05814 \times 1000 \ milli volts}$

$\mathsf{E_{Cl^-} \simeq - 58\ milli volts}$

5 0

3 years ago