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Alexeev081 [22]
3 years ago
14

Which of the following characteristics indicates that an atom is unstable?

Chemistry
2 answers:
qwelly [4]3 years ago
8 0

Answer:

D

Explanation:

Forming very strong bonds

ehidna [41]3 years ago
6 0

Answer:

I think is d.

Explanation:

particles that make up the nucleus are balanced. An atom is unstable (radioactive) if these forces are unbalanced; if the nucleus has an excess of internal energy. the Instability of an atom's nucleus may result from an excess of either neutrons or protons.

I'm not sure, but hope it helps.

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A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the
Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, C_6H_4N_2O_4

Explanation :

First we have to calculate the mass of benzene.

\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}

\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g  = 0.04395 kg

Formula used :  

\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}

where,

\Delta T_f = change in freezing point  = 5.53-1.37=4.16^oC

\Delta T_s = freezing point of solution

\Delta T^o = freezing point of benzene

Molal-freezing-point-depression constant (K_f) for benzene = 5.12^oC/m

m = molality

Now put all the given values in this formula, we get

4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}

\text{Molar mass of unknown compound}=180.6g/mol

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles

Moles of N = \frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{3.575}{1.193}=2.99\approx 3

For H = \frac{2.4}{1.193}=2.01\approx 2

For N = \frac{1.193}{1.193}=1

For O = \frac{2.381}{1.193}=1.99\approx 2

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = C_3H_2N_1O_2

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{180.6}{84}=2

Molecular formula = (C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4

Therefore, the molecular of the compound is, C_6H_4N_2O_4

3 0
3 years ago
The number 16 in carbon-16 stands for carbons
Oksanka [162]

Answer:

nice info

paragraph

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Explanation:

5 0
3 years ago
Fill in the following table.
Nostrana [21]

Answer:

i cant see the photo but i would love to help

Explanation:

6 0
2 years ago
In the laboratory, a volume of 100 mL of sulfuric acid (H2SO4) is recorded. How many g are there of the liquid if its density is
ser-zykov [4K]

Answer:

\large \boxed{\text{183 g}}  

Explanation:

\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\1.83 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{100 cm}^{3}}\\\\m & = & \text{183 g}\\\end{array}\\\text{There are $\large \boxed{\textbf{183 g}}$ of sulfuric acid.}

8 0
3 years ago
What is the capillary rise of ethanol in a glass tube with a 0.1 mm radius if the surface tension of ethanol is 0.032Jm2 and the
sweet [91]

Answer: The capillary rise(h) in the glass tube is = 0.009m

Explanation:

Using the equation

h = 2Tcosθ/rpg

Given

Contact angle, θ = Zero

h = height of the glass tube=?

T = surface tension = 0.032J/m^2

r = radius of the tube = 0.1mm =0.0001m

p= density of ethanol = 0.71g/cm^3

g= 9.8m/s^2

h = (2 * 0.032 * cos 0)/( 710*9.8*0.0001)

h= 0.09m

Therefore the capillary rise in the tube is 0.09m

5 0
3 years ago
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