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Morgarella
20 days ago
5

How many significant figures in 8400

Chemistry
2 answers:
castortr0y [2.7K]20 days ago
5 0
The number 8400 has two significant figures.
VMariaS [2.6K]20 days ago
4 0
There are two significant figures in the number 8400.
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In basic solution, se2− and so32− ions react spontaneously and e o cell = 0.35 v. (a) write the balanced half-reactions for this
lions [2649]

(a)   Write the balanced half-reactions for the overall process:

Oxidation: Se^2- (aq) → Se (s) + 2e-

Reduction: 2So3^2- (aq) + 3H2O (l) + 4e- → S2O3^2- + 6OH- (aq)

(b)   Assuming E sulfite is 0.57 V, compute E selenium:

E anode = E cathode – E cell

= -0.57 – 0.35

= -.092

3 0
21 day ago
The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C u
lions [2649]

Answer:

The enthalpy of the second intermediate equation is altered by halving its value and changing the sign.

Explanation:

Let's examine both the first and second intermediate reactions alongside the overall equation concerning the examined process;

First reaction;

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Thus, the overall reaction becomes;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH =?

According to Hess's law, which states that the total heat change in a reaction is equal to the sum of the heat changes for each step, we cannot simply sum the enthalpies for this overall reaction. Instead, we obtain the overall enthalpy by halving the second intermediate reaction's enthalpy and changing its sign before adding, as illustrated below;

Enthalpy of Intermediate reaction 1 + ½(-Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

7 0
1 month ago
Suppose a 20.0 g gold bar at 35.0°C absorbs 70.0 calories of heat energy. Given that the specific heat of gold is 0.0310 cal/g °
VMariaS [2690]

The change in temperature can be expressed as:

T_2-T_1=\dfrac{q}{mC_p(Gold)}

By substituting in the known values, we arrive at:

T_2-T_1=\dfrac{70\ cal}{20\ g\times 0.0310\ cal/g^o\ C}\\\\T_2-T_1=112.90^oC\\\\T_2-35^oC=112.90^oC\\\\T_2=(112.90+35)^oC\\\\T_2=147.9^oC

Thus, we obtain the required answer.

6 0
10 days ago
What is the density (in g/L) of a gas with a molar mass of 16.01 g/mol at 1.75 ATM and 337 K?
Anarel [2600]
To solve for density, you can use the formula--> Density= PM/ RT, where P stands for pressure, M for molar mass, R represents the gas constant, and T is temperature. 

P= 1.75 atm
M= 16.01 g/ mol
R= 0.0821 atm·L/ mol·K
T=337 k

Thus, the density calculation becomes: density= (1.75 x 16.01)/ (0.0821 x 337)= 1.01 g/L
8 0
9 days ago
Read 2 more answers
How much CO2 (L) is produced when 2.10 kg of sodium bicarbonate reacts with excess hydrochloric acid at 25.0 °C and 1.23 atm? A)
KiRa [2711]

The equation representing the reaction between sodium bicarbonate and hydrochloric acid is as follows:

NaHCO_3_(_s_) + HCl_(_a_q_) \implies NaCl_(_a_q_) + CO_2_(_g_) + H_2O_(_l_)

The substances NaHCO_3 and HCl combine in a 1:1 ratio. Therefore, we calculate the quantity of sodium bicarbonate and its molar mass to determine the moles formed.

NaHCO_3_M_r = 22.99 + 1.008 + 12.011+ 3 \times 16.0= 84.01 g/mol.

2.1kg\ NaHCO_3 \times \frac{1000g}{kg} \times \frac{mol}{84.01g/mol} = 24.997\ mol.

We also recognize that the stoichiometric proportions are 1:1:1:1:1, which leads to the conclusion that the moles of CO_2 equal 24.977 moles.

Next, we apply the ideal gas equation PV=nRT, where P denotes pressure, V refers to volume, R is the gas constant, and T represents the temperature in kelvins. We rearrange to solve for V

PV= nRT \implies V= \frac{nRT}{P}= \frac{ 24.997\ mol \times 8.2507m^3\ atm \times 298.15K }{mol \times K \times 1.23 atm} = 49967\ m^3

The final answer should be expressed in liters, 1L = 1000\ m^3, hence

49967\ m^3 \times\frac{L}{1000\ m^3} =49.97L\ CO_2\ produced

6 0
19 days ago
Read 2 more answers
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