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4 years ago

What does one mole of H2O correspond to?

1 answer:

6 0

the molar mass of H2O can be calculated with the following method:M(H2O)=2x M(H) + M(O), that is because we know that M(H)=1g/mol, M(O)=16g/mol, so M(H2O)=2x1+16=18g/mol, that means 1 mole of H2O contains 18g, so finally one mole of H2O correspond to 18 grams.

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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

This pluton occurs deep in Earth and does not cause any changes to the surface of Earth . True or False

irakobra [83]

Answer:

The given statement is false.

Explanation:

However, if the pluton exists beneath the ground, this could be conveniently shown in the illustration something from the peak such pluton appears convex in form resembling a lopolith and perhaps diapir, which would be a particular form of statistically significant pluton recognized as the sill.
Mostly from the figure it could also be shown that subsurface sheets are lined or curved, throughout the pluton mold. And therefore it is inferred that such a pluton creates adjustment to something like the ground atmosphere by altering the form of the levels above it.

So that the given is incorrect.

7 0

3 years ago

The molar mass of a compound is 92 g/mol. Analysis of a sample of the compound indicates that it contains 0.606g of N and 1.390g

Ghella [55]

Answer:

N2O4

Explanation:

To obtain the molecular formula of the compound, first, let us calculate the empirical formula for the compound. This is illustrated below:

N = 0.606g

O = 1.390g

Next, we divide the above by their molar masses

N = 0.606/14 = 0.0432

O = 1.390/16 = 0.0869

Next, we divide by the smallest

N = 0.0432/0.0432 = 1

O = 0.0869/0.0432 = 2

The empirical formula is NO2

The molecular formula is given by:

[NO2]n = 92

[14 + (16x2)]n = 92

[14 +32]n = 92

46n = 92

Divide both side by the coefficient of n i.e 46

n = 92/46

n = 2

The molecular formula = [NO2]n = [NO2]2 = N2O4

8 0

4 years ago

Read 2 more answers

Is it necessary to know the exact volume of the unknown acid or base to be titrated?

Darina [25.2K]

Yes, It is important to know the volume of Unknown acid or base to be titrated.

Titration is carried out in order to find out the concentration (i.e. molarity) of unknown acid or base. In this process a standard solution of acid or base is taken and is titrated with known volume of of titrant. At end point (neutralization) the amount of standard titrant utilized is calculated and following formula is employed to calculate the unknown concentration of unknown solution.

M₁V₁/n₁ = M₂V₂/n₂

7 0

3 years ago

Read 2 more answers

Of the following gases, ________ will have the greatest rate of effusion at a given temperature. Of the following gases, _______

kolezko [41]

Answer: From the given gases, the greatest rate of effusion is of $CH_4$

Explanation:

Rate of effusion of a gas is determined by a law known as Graham's Law.

This law states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$