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2 years ago

BRAINLIESTTT ASAP!!!

Chemistry

2 answers:

Stels [109]2 years ago

8 0

While I am not the brainliest I can certainly answer.

This was a chemical change because the chemical components were changed, a big giveaway to this was the fizzing, however the temperature rising was also another giveaway.

Vladimir79 [104]2 years ago

4 0

Explanation:

A physical change is the change that does not bring any difference in the chemical properties of a substance.

For example, change in shape, size, state, mass, density etc are all physical changes.

On the other hand, a chemical change is define as the one which brings difference in chemical composition of a substance.

For example, yeast granules were added to hydrogen peroxide then fizzing and bubbling took place. This means that a chemical reaction has taken place due to which a gas is liberating out of the solution.

Also, heat is liberating due to which temperature has began to rise.

Thus, we can conclude that based on the given observation a chemical change has taken place.

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A gas balloon has a volume of 80.0 mL at 300K , and a pressure of 50.0 kPa. if the pressure changes to 80.0 kPa and the temperat

stellarik [79]

Answer: 53.3

Explanation:

V2=(T2 x P1 x V1)/(T1 x P2)

(320x50x80)/(300x80)

53.3

3 0

2 years ago

The gas in a sealed container has an absolute pressure of 125.4 kilopascals. If the air around the container is at a pressure of

AlexFokin [52]

Answer: C. 25.6 kPa

Explanation:

The Gauge pressure is defined as the amount of pressure in a fluid that exceeds the amount of pressure in the atmosphere.

As such, the formula will be,

PG = PT – PA

Where,

PG is Gauge Pressure

PT is Absolute Pressure

PA is Atmospheric Pressure

Inputted in the formula,

PG = 125.4 - 99.8

PG = 25.6 kPa

The gauge pressure inside the container is 25.6kPa which is option C.

4 0

2 years ago

If 25 g of NH3, and 96 g of H2S react according to the following reaction, what is the

jeyben [28]

25 g of NH₃ will produce 47.8 g of (NH₄)₂S

<u>Explanation:</u>

2 NH₃ + H₂S ----> (NH₄)₂S

Molecular weight of NH₃ = 17 g/mol

Molecular weight of (NH₄)₂S = 68 g/mol

According to the balanced reaction:

2 X 17 g of NH₃ produces 68 g of (NH₄)₂S

1 g of NH₃ will produce $\frac{68}{34}$ g of (NH₄)₂S

25g of NH₃ will produce $\frac{65}{34} X 25 g$ of (NH₄)₂S

= 47.8 g of (NH₄)₂S

Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S

4 0

2 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

2 years ago

What is the molar mass of al (clo2)3

frozen [14]

The molar mass is 229.33

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2 years ago