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goblinko [34]
3 years ago
11

Show how the total charge sodium oxide is zero

Chemistry
1 answer:
coldgirl [10]3 years ago
6 0
First, the symbol for sodium oxide is Na₂O

Each Na (sodium) has a charge of 1+, and each O has a charge of 2-  :

Na₂¹⁺O²⁻

There are two Na's, however, and each one is 1+, however, so the Na₂ has a total charge of 2+.  Because of this, the 2+ from the 2 Na's and the 2- from the O cancel each other out to make 0.
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Please I need help with all the questions 1-5 and inn struggling with it and if you need to see the picture big then click on it
irakobra [83]
It’s hard to zoom in and read the questions
8 0
4 years ago
Para formar bronce, se mezclan 150g de cobre a 1100°C y 35g de estaño a 560°C. Determine la temperatura final del sistema.
jek_recluse [69]

Answer:

La temperatura final del sistema es 1029,346 °C.

Explanation:

Asumamos que el sistema conformado por el cobre y el estaño no tiene interacciones con sus alrededores. Por la Primera Ley de la Termodinámica, el cobre cede calor al estaño con tal de alcanzar el equilibrio térmico. El cobre se encuentra inicialmente en su punto de fusión, mientras que el estaño está por encima de ese punto, de modo que la transferencia de calor es esencialmente sensible:

m_{Cu}\cdot c_{Cu}\cdot (T-T_{Cu}) = m_{Sn}\cdot c_{Sn}\cdot (T_{Sn}-T)

(m_{Cu}\cdot c_{Cu} + m_{Sn}\cdot c_{Sn})\cdot T = m_{Sn}\cdot c_{Sn}\cdot T_{Sn} + m_{Cu}\cdot c_{Cu}\cdot T_{Cu}

T = \frac{m_{Sn}\cdot c_{Sn}\cdot T_{Sn}+m_{Cu}\cdot c_{Cu}\cdot T_{Cu}}{m_{Cu}\cdot c_{Cu}+m_{Sn}\cdot c_{Sn}} (1)

Donde:

m_{Sn} - Masa del estaño, en gramos.

m_{Cu} - Masa del cobre, en gramos.

c_{Sn} - Calor específico del estaño, en calorías por gramo-grados Celsius.

c_{Cu} - Calor específico del cobre, en calorías por gramo-grados Celsius.

T_{Sn} - Temperatura inicial del estaño, en grados Celsius.

T_{Cu} - Temperatura inicial del cobre, en grados Celsius.

Si sabemos que m_{Cu} = 150\,g, m_{Sn} = 35\,g, c_{Cu} = 0,093\,\frac{cal}{g\cdot ^{\circ}C}, c_{Sn} = 0,060\,\frac{cal}{g\cdot ^{\circ}C}, T_{Sn} = 560\,^{\circ}C y T_{Cu} = 1100\,^{\circ}C, entonces la temperatura final del sistema es:

T = \frac{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (560\,^{\circ}C)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (1100\,^{\circ}C)}{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)}

T = 1029,346\,^{\circ}C

La temperatura final del sistema es 1029,346 °C.

3 0
3 years ago
What is the pH of a solution that is 0.40 M NaBrO and 0.50 M HBrO (hypobromous acid) (Ka for HBrO = 2.3 x 10^-9)
Pie

Answer

pH=8.5414

Procedure

The Henderson–Hasselbalch equation relates the pH of a chemical solution of a weak acid to the numerical value of the acid dissociation constant, Kₐ. In this equation, [HA] and [A⁻] refer to the equilibrium concentrations of the conjugate acid-base pair used to create the buffer solution.

pH = pKa + log₁₀ ([A⁻] / [HA])

Where

pH = acidity of a buffer solution

pKa = negative logarithm of Ka

Ka =acid disassociation constant

[HA]= concentration of an acid

[A⁻]= concentration of conjugate base

First, calculate the pKa

pKa=-log₁₀(Ka)= 8.6383

Then use the equation to get the pH (in this case the acid is HBrO)

pH=8.6383+\log_{10}(\frac{0.40\text{ M}}{0.50\text{ M}})=8.5414

8 0
1 year ago
What’s the answer for number 2?
Lady bird [3.3K]
The answer to this is surface wave i think
4 0
4 years ago
Read 2 more answers
How many moles of H2O2 are needed to react with 1.07 moles of N2H4?
I am Lyosha [343]

Answer:

2.14 moles of H₂O₂ are required

Explanation:

Given data:

Number of moles of H₂O₂ required = ?

Number of moles of N₂H₄ available = 1.07 mol

Solution:

Chemical equation:

N₂H₄  +   2H₂O₂       →   N₂ +  4H₂O

now we will compare the moles of H₂O₂ and N₂H₄

                          N₂H₄     :      H₂O₂  

                            1           :        2

                            1.07      :         2×1.07 = 2.14 mol

                   

6 0
3 years ago
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