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3 years ago

Where are the most reactive nonmetal elements found on the periodic table?

Chemistry

2 answers:

xz_007 [3.2K]3 years ago

7 0

A. The most reactive Non-metals are found at the top of Group 17 (Halogens) in the form of Fluorine and Chlorine

Marina86 [1]3 years ago

3 0

Answer:

THE ANSWER IS at the top of group 17

Explanation:

Your welcome~! c:

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The point on a magnet Where magnet effect is strongest is called

notsponge [240]

Pole

Explanation:

A magnetic pole is the point on a magnet where a magnet effect is strongest. Around a magnet, there are magnetic fields.

The magnetic fields are force fields that leaves the north poles and enters through the south.
The pole is where magnetic effects are the strongest.
It is around the pole that the strongest magnetic effect on a magnet or magnetic material is felt.
The strength of magnetic fields are strongest at the poles.

learn more:

electromagnet brainly.com/question/2191993

#learnwithBrainly

5 0

3 years ago

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

Determine the molarity of 56.00 grams of aluminum nitrate in 450 milliliters of solution?

SVETLANKA909090 [29]

Answer:

4.62 M

Explanation:

Molarity = moles/volumes (L), so you need to find the moles and the volumes in liters.

Finding the volume is easy because you just have to convert mL to L, so the volume is 0.45 L

Next, find the moles. You can do this by using the molar mass of aluminum to convert the grams to moles. The molar mass of aluminum is 26.98 g/mol.

56 g * (1 mol/26.98 g) = 2.08 mol

Now, divide the moles (2.08) by the volume (.45 L)

Molarity = 4.62 M

3 0

3 years ago

A 75.0 g sample of dinitrogen monoxide is confined in a 3.q L vessel. What is the pressure( in atm) at 115 celsius

Nonamiya [84]

Data Given:
Pressure = P = ?

Volume = V = 3.0 L

Temperature = T = 115 °C + 273 = 388 K

Mass = m = 75.0 g

M.mass = M = 44 g/mol

Solution:
Let suppose the Gas is acting Ideally. Then According to Ideal Gas Equation,
P V = n R T
Solving for P,
P = n R T / V ------ (1)
Calculating Moles,
n = m / M

n = 75.0 g / 44 g.mol⁻¹

n = 1.704 mol

Putting Values in Eq. 1,

P = (1.704 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 388 K) ÷ 3.0 L

P = 18.08 atm

7 0

3 years ago

The table below shows the height of a ball x seconds after being kicked.

irga5000 [103]

Answer:

-16

Explanation:

8 0

2 years ago

Read 2 more answers