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Degger [83]
3 years ago
11

Select the correct answer. The gas in a sealed container has an absolute pressure of 9.25 atmospheres. If the air around the con

tainer is at standard pressure, what is the gauge pressure inside the container? A. 0.759 atm B. 8.25 atm C. 10.25 atm D. 113 atm

Chemistry
2 answers:
N76 [4]3 years ago
7 0

Answer:

Explanation:

8.25 atm

yuradex [85]3 years ago
4 0

Answer: B) 8.25 atm.

Explanation: Absolute pressure is the sum of gauge pressure and atmospheric pressure.

P_a_b_s_o_l_u_t_e=P_g_a_u_g_e+P_a_t_m_o_s_p_h_e_r_e

Atmosphere pressure is given as 9.25 atm and the atmospheric pressure is 1.00 atm.

Let's plug in the values in the above formula:

9.25 atm = P_g_a_u_g_e + 1.00 atm

P_g_a_u_g_e = 9.25 atm - 1.00 atm

P_g_a_u_g_e = 8.25 atm

So, gauge pressure inside the container is 8,25 atm and hence the right option is B.

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Hello, what’s the difference between oxidation and oxidising agents?
lions [1.4K]

Answer:

An oxidising agent oxidises something else. Oxidation is loss of electrons (OIL RIG). That means that an oxidising agent takes electrons from that other substance.

Explanation:

7 0
2 years ago
Consider a balloon with volume V. It contains n moles of gas and has an internal pressure of P. The temperature of the gas is T.
Maksim231197 [3]

Answer:

V₂ = 0.6 V.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If n is constant, and have different values of P, V and T:

<em>(P₁V₁T₂) = (P₂V₂T₁).</em>

<em></em>

V₁ = V, P₁ = P, T₁ = T.

V₂ = ??? V, ​P₂ = 1.25 P, T₂ = 0.75 T.

<em>∴ V₂ = (P₁V₁T₂)/(P₂T₁) =</em> (P)(V)(0.75 T)/(1.25 P)(T)<em> = 0.6 V.</em>

3 0
3 years ago
Read 2 more answers
The mass of solute per 100 mL of solution is abbreviated as (m/v). Mass is not technically the same thing as weight, but the abb
Nimfa-mama [501]

Answer:

\boxed{\text{254 g}}

Explanation:

\begin{array}{rcl}\text{\% m/V} & = & \dfrac{\text{Mass of sucrose}}{\text{Volume of solution}}\\\\\text{Let m}& = &\text{mass of sucrose}\\\dfrac{\text{35.0 g}}{\text{100 mL}}& = & \dfrac{m}{\text{725 mL}}\\\\m & = &\dfrac{\text{35.0 g}\times 725}{100}\\\\ & = &\textbf{254 g}\\\end{array}\\\text{You need $\boxed{\textbf{254 g}}$ of sucrose}

3 0
3 years ago
If the solubility of KCl in 100 mL of H₂O is 34 g at 20 °C and 43 g at 50 °C, label each of the following solutions as unsaturat
sertanlavr [38]

Answer:

a) Unsaturated

b) Supersaturated

c) Unsaturated

Explanation:

A saturated  solution contains the <u>maximum amount of a solute that will dissolve in a given  solvent at a specific temperature</u>.

An unsaturated solution contains <u>less solute than it  has the capacity to dissolve. </u>

A supersaturated solution, <u>contains more  solute than is present in a saturated solution</u>. Supersaturated solutions are not very  stable. In time, some of the solute will come out of a supersaturated solution as crystals.

According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at <u>20 °C is 34 g</u>, and at <u>50 °C is 43 g</u> we can label the solutions:

a) 30 g in 100 mL of H₂O at 20 °C  ⇒ unsaturated

b) 65 g in 100 mL of H₂O at 50 °C  ⇒ supersaturated

c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution <u>with no precipitate</u> ⇒ unsaturated (if it were saturated it would have had precipitate)

8 0
3 years ago
PLEASE HELP ME WITH MY HOMEWORK!!!!
vovangra [49]
3.gases- <span> Their molecules are already the farthest apart compared with solids and liquids, and small changes in temperature causes these loose molecules.</span>
4 0
3 years ago
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