What is the mass in grams of 1.204 x 1024 molecules of

Which of these relationships is true of an uncharged atom?

marusya05 [52]

Answer:

The correct option is C) The number of protons is equal to the number of electrons.

Explanation:

Atoms are made up of particles called protons, neutrons, and electrons. Protons and neutrons are in the center of the atom and form the nucleus, while electrons surround the nucleus. Electrons have a negative charge. The charge of the protons is positive and finally, the neutrons have no charge.

If the atom has no charge, this means that the total charge of the atomic nucleus, which is positive due to the presence of the protons, is equal to the negative charge of the electrons, so that it cancels out.

So, <u><em>the correct option is C) The number of protons is equal to the number of electrons.</em></u>

7 0

3 years ago

Calculate the concentration of the resulting sodium oxalate solution if 0.1005 g of Na2C2O4 was used.

sashaice [31]

Answer:

0.00500M of Na₂C₂O₄

Explanation:

<em>When are dissolved in 150 mL of 1.0 M H2SO4.</em>

<em />

We can solve this problem finding molarity of sodium oxalate: That is, moles of Na2C2O4 per liter of solution. Thus, we need to convert the 0.1005g to moles using molar mass of sodium oxalate (134g/mol) and dividing in the 0.150L of the solution:

0.1005g * (1mol / 134g) = 7.5x10⁻⁴ moles of Na₂C₂O₄

In 0.150L:

7.5x10⁻⁴ moles of Na₂C₂O₄ / 0.150L =

4 0

3 years ago

What was the biggest challenge you faced in getting to where you are today and how did you overcome it?

Alchen [17]

It’s all based on you but yours could be anything hard you went through and how you overcame it. Also say how it made you the person you are today

4 0

3 years ago

2KI + Pb(NO3)2 → 2KNO3 + PbI2 Determine how many moles of KNO3 are created if 0.03 moles of KI are completely consumed.

ElenaW [278]

2:2 so the same proportion
0,03

6 0

3 years ago

Read 2 more answers

determine the empirical and molecular formula of a compound composed of 18.24 g carbon, 0.51 g hydrogen, and 16.91 g fluorine an

Vika [28.1K]

Answer: The empirical formula for the given compound is $C_3HF_2$ and molecular formula for the given compound is $C_{24}H_8F_{16}$

Explanation : Given,

Mass of C = 18.24 g

Mass of H = 0.51 g

Mass of F = 16.91 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{18.24g}{12g/mole}=1.52moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.51g}{1g/mole}=0.51moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{16.91g}{19g/mole}=0.89moles$

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 0.51 moles.

For Carbon = $\frac{1.52}{0.51}=2.98\approx 3$

For Hydrogen = $\frac{0.51}{0.51}=1$

For Fluorine = $\frac{0.89}{0.51}=1.74\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : F = 3 : 1 : 2

The empirical formula for the given compound is $C_3H_1F_2=C_3HF_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Molar mass = 562.0 g/mol

Mass of empirical formula = 3(12) + 1(1) + 2(19) = 75 g/eq

Putting values in above equation, we get:

$n=\frac{562.0}{75}=7.49\approx 8$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_3HF_2=(C_3HF_2)_n=(C_3HF_2)_8=C_{24}H_8F_{16}$

Thus, the molecular formula for the given compound is $C_{24}H_8F_{16}$

8 0

3 years ago