What is the mass in pounds of 1.000 quart of mercury

Calculate the amount of oxygen gas collected by the displacement of water at 14◦C if the atmospheric pressure is 790 Torr and th

zhannawk [14.2K]

Answer : The amount of oxygen gas collected are, 0.217 mol

Explanation :

Using ideal gas equation :

$PV=nRT$

where,

P = pressure of gas = $(790-12)torr=778torr=1.02atm$ (1 atm = 760 torr)

V = volume of gas = 5 L

T = temperature of gas = $14^oC=273+14=287K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1.02atm)\times (5L)=n\times (0.0821L.atm/mol.K)\times (287K)$

$n=0.217mole$

Thus, the amount of oxygen gas collected are, 0.217 mol

8 0

4 years ago

andreyandreev [35.5K]

Answer:

H3PO4 is stronger than H2PO4- because

H3PO4 dissociation constant is 6.9×10^-3

H2PO4^- dissociation constant is 6.2×10^-8

7 0

2 years ago

Chris noticed a high-pressure system in the weather forecast for Monday

Alchen [17]

Answer:

A. Clear and Sunny

Explanation:

The answer would be A. Clear and Sunny. A high pressure system occurs where the air mass above the Earth is denser than in surrounding areas, and therefore exerts a higher force or pressure. They usually happen with li9ght winds. Using process of elimination, it could not be D. Warm and Stormy, C. Cloudy and Rainy, or B. Cold and Stormy because all have wet climates.

5 0

2 years ago

chlorine has 7 outer energy level electrons and argon has 8. what is the difference in the reactivity of these two elements?

Lunna [17]

Chlorine is more reactive than argon, because argon is not reactive at all. Chlorine only has to gain 1 valence electron

6 0

3 years ago

Read 2 more answers

A 44.0 g sample of an unknown metal at 99.0 oC was placed in a constant-pressure calorimeter of negligible heat capacity contain

tatiyna

Answer:

$C_m=0.474\frac{J}{g\°C}$

Explanation:

Hello.

In this case, since this is a system in which the water is heated up and the metal is cooled down in a calorimeter which is not affected by the heat lose-gain process, we can infer that the heat lost by the metal is gained be water, it means that we can write:

$Q_m=-Q_w$

Thus, in terms of masses, specific heats and temperatures we can write:

$m_mC_m(T_{eq}-T_m)=-m_wC_w(T_{eq}-T_w)$

Whereas the equilibrium temperature is the given final temperature of 28.4 °C and we can compute the specific heat of the metal as shown below:

$C_m=\frac{-m_wC_w(T_{eq}-T_w)}{m_m(T_{eq}-T_m)}$

Plugging the values in and since the density of water is 1.00 g/mL so the mass is 80.0g, we obtain:

$C_m=\frac{-80.0g*4.184\frac{J}{g\°C} (28.4\°C-24.0\°C)}{44.0g(28.4\°C-99.0\°C)}\\\\C_m=0.474\frac{J}{g\°C}$

Best regards!

6 0

3 years ago