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2 years ago

Plsssssss! help! 20 points

Chemistry

1 answer:

Evgen [1.6K]2 years ago

5 0

Answer:

Explanation:

molecular weight of N₂H₄ is as follows

2 x 14 + 4 x 1 = 32

so 32 gram of N₂H₄ is equal to one mole of N₂H₄

25 gram of N₂H₄ will be equal to 25 / 32 mole of N₂H₄

so numerical quantity needed to convert grams of N₂H₄ to mole is

1 / 32 or .03125 . This number needs to be multiplied with grams of N₂H₄

to convert it to number of moles.

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Describe an experience you've had making or building something where the amount of each ingredient or building block came in fix

nignag [31]

Baking a cake is an example of making something where the ingredients must be in fixed ratios. Recipes call for specific ratios of ingredients in order to cook properly, and when a recipe for a cake is modified to feed greater or fewer people the ratio remains the same as the original recipe.

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2 years ago

Aluminum–lithium (Al–Li) alloys have been developed by the aircraft industry to reduce the weight and improve the performance of

gtnhenbr [62]

Answer:

The concentration of Li (in wt%) is 3,47g/mol

Explanation:

To obtain the 2,42g/cm³ of density:

2,42g/cm³ = 2,71g/cm³X + 0,534g/cm³Y (1)

Where X is molar fraction of Al and Y is molar fraction of Li.

X + Y = 1 (2)

Replacing (2) in (1):

Y = 0,13

Thus, X = 0,87

The weight of Al and Li is:

0,87*26,98g/mol = 23,4726 g of aluminium

0,13*6,941g/mol = 0,84383 g of lithium

The concentration of Li (in wt%) is:

0,84383g/(0,84383g+23,4726g) ×100= 3,47%

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1 year ago

The decomposition of nitramide, O 2 NNH 2 , O2NNH2, in water has the chemical equation and rate law O 2 NNH 2 ( aq ) ⟶ N 2 O ( g

valkas [14]

Answer:

Explanation:

The given overall reaction is as follows:

O 2 N N H ₂( a q ) k → N ₂O ( g ) + H ₂ O( l )

The reaction mechanism for this reaction is as follows:

O ₂ N N H ₂ ⇌ k 1 k − 1 O ₂N N H ⁻ + H ⁺ ( f a s t e q u i l i b r i u m )

O ₂ N N H − k ₂→ N ₂ O + O H ⁻ ( s l ow )

H ⁺ + O H − k ₃→ H ₂ O ( f a s t )

The rate law of the reaction is given as follows:

k = [ O ₂ N N H ₂ ] / [ H ⁺ ]

The rate law can be determined by the slow step of the mechanism.

r a t e = k ₂ [ O ₂ N N H ⁻ ] . . . ( 1 )

Since, from the equilibrium reaction

k e q = [ O ₂ N N H ⁻ ] [ H ⁺ ] /[ O ₂ N N H ₂ ] = k ₁ /k − 1

[ O ₂ N N H ⁻] = k ₁ /k − 1 × [ O ₂ N N H ₂ ] /[ H ⁺ ]. . . . ( 2 )

Substitituting the value of equation (2) in equation (1) we get.

r a t e = k ₂ k ₁/ k − 1 × [ O ₂ N N H ₂ ] /[ H ⁺ ]

Therefore, the overall rate constant is

k = k₂k₁/k-1

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1 year 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

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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 $3.4\times 10\; \rm atm$ .

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allochka39001 [22]

Explanation:

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