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Nikolay [14]
3 years ago
11

Name one chemical property of carbon disulfide

Chemistry
1 answer:
butalik [34]3 years ago
3 0
One carbon and 2 sulfurs
You might be interested in
Give the empirical formula for C↓8H↓8
mr_godi [17]

Answer:

Empirical formula of  C₈H₈ =  CH

Explanation:

Data Given:

Molecular Formula = C₈H₈

Empirical Formula = ?

Solution

Empirical Formula:

Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.

So,

tha ration of the molecular formula should be divided by whole number to get the simplest ratio of molecule

        C₈H₈ Consist of  Carbon (C), and Hydrogen (H)

Now

Look at the ratio of these two atoms in the compound

                         C : H

                         8 : 8

Divide the ratio by two to get simplest ratio

                          C : H

                       8/8 : 8/8

                         1   :  1

So for the empirical formula is the simplest ratio of carbon to hydrogen 1 : 1

So the empirical formula will be

                     Empirical formula of  C₈H₈ =  CH

4 0
3 years ago
How many are molecules ( or formula) in each sample?
andre [41]

Answer:

  • 4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}
  • 16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

<u>Explanation</u>:

<u>Number of molecules for 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}</u>

\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:

Atomic mass of Na + H + C + 3(O)  = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }

55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)

=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}

<u>Number of molecules for for \left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}</u>

\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.

=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

8 0
3 years ago
As an object falls down the earth, its gravitational potential energy will
Travka [436]

Answer:

a. decrease

Explanation:

6 0
2 years ago
Given the following equation: Cu + 2 AgNO3 ---&gt; Cu(NO3)2 + 2
Temka [501]

Given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.

<h3>How to calculate mass of substances?</h3>

The mass of a substance can be calculated using the following steps:

Cu + 2AgNO3 = Cu(NO3)2 + 2Ag

1 mole of Cu react with 2 moles of AgNO3

  • Molar mass of AgNO3 = 169.87 g/mol
  • Molar mass of Cu = 63.5g/mol

moles of AgNO3 = 262g/169.87g/mol = 1.54mol

1.54 moles of AgNO3 will react with 0.77 moles of Cu.

mass of Cu = 0.77 × 63.5 = 48.97g

Therefore, given the following equation; Cu + 2AgNO3 = Cu(NO3)2 + 2Ag, 48.97 grams of Cu are needed to react with 262g of AgNO3.

Learn more about mass at: brainly.com/question/6876669

8 0
2 years ago
A city's water supply is contaminated with a toxin at a concentration of 0.63 mg/L. For the water to be safe for drinking, the c
Shalnov [3]

Answer:

Approximately 22.37 days, will it take for the water to be safe to drink.

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

k is rate constant

Given that:- k = 0.27 (day)⁻¹

[A_0] = 0.63 mg/L

[A_t]=1.5\times 10^{-3} mg/L

Applying in the above equation as:-

1.5\times 10^{-3}=0.63e^{-0.27\times t}

63e^{-0.27t}=150\times \:10^{-3}

e^{-0.27t}=\frac{1}{420}

t=\frac{100\ln \left(420\right)}{27}=22.37

<u>Approximately 22.37 days, will it take for the water to be safe to drink.</u>

7 0
3 years ago
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