All categories

3 years ago

In a short paragraph, explain one thing you've learned from the course this year.

Chemistry

1 answer:

exis [7]3 years ago

7 0

Answer:

ive learned about the chart , ive learned about the coolest chemicle reactions and how. the can change colors , ive also learned that chenistrey is all around us since the beginning of life ,and how liquid can turn solid

Explanation:

You might be interested in

B. from 52 moles of nitric acid, how many moles of silver nitrate will be produced?

IceJOKER [234]

Answer:

39 mol AgNO3

Explanation:

We have the equation 4HNO3 + 3Ag -----> 3AgNO3 + NO + 2H2O

We want to calculate the number of silver nitrate (AgNO3) moles that would be produced from 52 moles of nitric acid ( HNO3 )

We can calculate this by using mole ratio as well as dimensional analysis.

The mole ratio of Silver nitrate to nitric acid based on the balanced equation is 3AgNO3:4HNO3.

Using this we can create a table: The table is attached.

Breakdown of the table.

The moles of nitric acid cancel out and we multiply 52 by 3/4 to get 39 moles of Silver nitrate.

6 0

2 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

How many grams of NH3 can be produced from 2.30 mol of N2 and excess H2.

MrRa [10]

<h3>Answer:</h3>

78.34 g

<h3>Explanation:</h3>

From the question we are given;

Moles of Nitrogen gas as 2.3 moles

we are required to calculate the mass of NH₃ that may be reproduced.

<h3>Step 1: Writing the balanced equation for the reaction </h3>

The Balanced equation for the reaction is;

N₂(g) + 3H₂(g) → 2NH₃(g)

<h3>Step 2: Calculating the number of moles of NH₃</h3>

From the equation 1 mole of nitrogen gas reacts to produce 2 moles of NH₃

Therefore, the mole ratio of N₂ to NH₃ is 1 : 2

Thus, Moles of NH₃ = Moles of N₂ × 2

= 2.3 moles × 2

= 4.6 moles

<h3>Step 3: Calculating the mass of ammonia produced </h3>

Mass = Moles × molar mass

Molar mass of ammonia gas = 17.031 g/mol

Therefore;

Mass = 4.6 moles × 17.031 g/mol

= 78.3426 g

= 78.34 g

Thus, the mass of NH₃ produced is 78.34 g

3 0

3 years ago

Which of the following statements is not correct?

Lilit [14]

Answer:

It's 4.1 moles Li2O = 94 g on Odyssey ware

8 0

2 years ago

Read 2 more answers

Convert 6.33×10−6 cg to nanograms.

tensa zangetsu [6.8K]

5.73e+8 hope this help

5 0

3 years ago