Ask question

Ask question

All categories

3 years ago

6

A tire at 21°C has a pressure of 0.82 atm. Its temperature decreases to –3.5°C. If there is no volume change in the tire, what i

s the pressure after the temperature change?

1 answer:

Shtirlitz [24]3 years ago

5 0

Answer:

a

Explanation:

0.75

You might be interested in

A 1.11 g sample of caffeine, C8H10N4O2, burns in a constant-volume calorimeter that has a heat capacity of 7.85 kJ/K. The temper

VladimirAG [237]

Answer:

401135 kJ

Explanation:

From the balanced quation,

(q/n) = CΔE

Molar heat of combustion = 7.85kJk × (303.81-298.70)k

= 7.85kj × 5.11

= 40.1135kj

4 0

2 years ago

A 1.555-g sample of baking soda decomposes with heat to produce 0.991 g Na2CO3. Refer to Example Exercise 14.l and show the calc

Sunny_sXe [5.5K]

Answer:

a) 101%

b)59.7%

Explanation:

The equation for the thermal decomposition of baking soda is shown;

2NaHCO3 → Na2CO3 + H2O + CO2

Number of moles of baking soda= mass/molar mass= 1.555g/84.007 g/mol = 0.0185 moles

From the reaction equation;

2 moles of baking soda yields 1 mole of sodium carbonate

0.0185 moles of baking soda will yield = 0.0185 moles ×1 /2 = 9.25 ×10^-3 moles of sodium carbonate.

Therefore, mass of sodium carbonate= 9.25 ×10^-3 moles × 106gmol-1= 0.9805 g of sodium carbonate. This is the theoretical yield of sodium carbonate.

%yield = actual yield/theoretical yield ×100

% yield = 0.991/0.9805 ×100

%yield = 101%

Since ;

2NaHCO3 → Na2CO3 + H2O + CO2

And H2O + CO2 ---> H2CO3

Hence I can write, 2NaHCO3 → Na2CO3 + H2CO3

Molar mass of H2CO3= 62.03 gmol-1

Molar mass of baking soda= 84 gmol-1

Therefore, mass of baking soda=

0.325/62.03 × 2 × 84 = 0.88 g of NaHCO3

% of NaHCO3= 0.88/1.473 × 100 = 59.7%

7 0

2 years ago

Al(s) + NiSO4(aq)<br> Balance

irinina [24]

Answer:

$2Al(s) + 3NiSO4 --> Al2(SO4)3 + 3Ni$

Explanation:

This reaction type is a single replacement. The format of a single replacement is:

$A+BC --> B+AC$

A= Al

B= Ni

C= SO

The coefficient 3 for Ni would become a subscript for AC. After you plug those into the reaction you need to count how many of each are on the left side and try to get the same number on the right side. Both sides must be equal to have a balanced equation.

8 0

3 years ago

A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

musickatia [10]

Answer:

$T_2=335.42K=62.27^oC$

Explanation:

Hello,

In this case, by using the general gas law, that allows us to understand the pressure-volume-temperature relationship as shown below:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

Thus, solving for the temperature at the end (considering absolute units of Kelvin), we obtain:

$T_2=\frac{P_2V_2T_1}{P_1V_1}=\frac{1.8L*0.75atm*(25+273.15)K}{1.2L*1.0atm} \\\\T_2=335.42K=62.27^oC$

Best regards.

4 0

3 years ago

Read 2 more answers

Hemoglobin (Hb) can be viewed as having two quaternary states, a low oxygen affinity state (T), and a high oxygen affinity state

natta225 [31]

<h3><u>Answer;</u></h3>

The statements that are True are;

Upon binding a molecule of oxygen, Hb undergoes a conformational change that makes the binding of subsequent O2 molecules easier.
The conformational change induced in Hb upon binding oxygen is the result of a small movement (0.2 Å) of the iron cation in the center of heme.
Site-directed mutagenesis studies have indicated that the cooperativity of O2 binding in Hb is attributable to the movement of the F helix in Hb.

<h3><u>Explanation</u>;</h3>

Hemoglobin is a key pigment in the blood that transports oxygen gas to all the tissues in the body. It is made up of two types of chains; that is two alpha chains and two beta chains.
in its deoxygenated state hemoglobin has a low affinity for oxygen compared to myoglobin. When oxygen is bound to the first subunit of hemoglobin it leads to subtle changes to the quaternary structure of the protein. This in turn makes it easier for a subsequent molecule of oxygen to bind to the next subunit.

8 0

3 years ago