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Oksi-84 [34.3K]
2 years ago
14

Can you help me with three please? We’re balancing electrons

Chemistry
1 answer:
Aneli [31]2 years ago
4 0

Answer:

Explanation:

I did this class yesterday give me like 10min imma find my anwsers

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When a sample is heated water is observed at the top of the tube. The compound melts and changes color. Te residue is not entire
harkovskaia [24]

Answer:

hydrate

Explanation:

when a hydrate is heated,it changes color due to the exothermic reaction taking place.the structure of the complex changes but not entirely.this result in the sample to to not dissolve completely and we can observe the small traces of the sample.

7 0
3 years ago
Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o
Gnom [1K]

Answer: The empirical formula for the given compound is C_5H_7N

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:

C_xH_yN_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.

We are given:

Mass of CO_2=21.363mg=21.363\times 10^3g=21363g

Mass of H_2O=6.125g=6.125\times 10^3g=6125g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 21363 g of carbon dioxide, \frac{12}{44}\times 21363=5826.27g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, \frac{2}{18}\times 6125=680.55 of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = \frac{485.52}{97.73}=4.96\approx 5

For Hydrogen  = \frac{680.55}{97.73}=6.96\approx 7

For Nitrogen = \frac{97.73}{97.73}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is C_5H_7N_1=C_5H_7N

7 0
3 years ago
A student is asked to convert measurements from liters to milliliters and from milliliters to liters. His answers are shown in b
kirza4 [7]
The answer is 3.6L=360mL
5 0
3 years ago
Read 2 more answers
Define exothermic reaction.​
Maksim231197 [3]

An exothermic reaction is a reaction that releases energy.

7 0
3 years ago
Read 2 more answers
Se tiene una solución acuosa 2M de carbonato de potasio. Expresar su concentración en %p/v y Normalidad.
zepelin [54]

Answer:

Normalidad = 4N

%p/V = 27.6%

Explanation:

La solución 2M de carbonato de potasio contiene 2moles de carbonato por litro de solución. La normalidad son los equivalente de carbonato de potasio (2eq/mol) por litro de solución:

2moles * (2eq/mol) = 4eq / 1L = 4N

El porcentaje peso volumen es el peso de carbonato en gramos dividido en el volumen en mL por 100:

%p/V:

Masa K2CO3 -Masa molar: 138.205g/mol-

2moles * (138.205g/mol) = 276g K2CO3

Volumen:

1L * (1000mL/1L) = 1000mL

%p/V:

276g K2CO3 / 1000mL * 100

<h3>%p/V = 27.6%</h3>
6 0
3 years ago
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