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3 years ago

Write a balanced chemical equation for the neutralization reaction between phosphoric acid and rubidium hydroxide.

1 answer:

5 0

Phosphoric acid is a weak acid, while rubidium hydroxide is a strong acid.

H₃PO₄ + RbOH --> Rb₃PO₄ + H₂O

We get Rb₃PO₄ because PO₄ has a charge of 3-, that is PO₄³⁻. Rb has a charge of 1+. You give the subscript of one the charge of the other as this is an ionic compound. So you end up with Rb₃PO₄, a neutral compound.

Now let's balance the equation:
H₃PO₄ + 3RbOH --> Rb₃PO₄ + 3H₂O

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What are the 10 element

maw [93]

hydrogen
oxygen
carbon
nitrogen
fluorine
silicone
boron
argon
cobalt
aluminum

hope it hepls✨

4 0

3 years ago

The empirical formula of a compound is CH2O and its molecular weight is 90.09 g/mol. Determine the molecular formula.

Pachacha [2.7K]

Answer : The molecular formula of the compound will be, $C_{3}H_{6}O_3$

Explanation :

Empirical formula : It is the simplest form of the chemical formula which depicts the whole number of atoms of each element present in the compound.

Molecular formula : it is the chemical formula which depicts the actual number of atoms of each element present in the compound.

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 :

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

As we are given that the empirical formula of a compound is $CH_2O$ and the molar mass of compound is, 90.09 gram/mol.

The empirical mass of $CH_2O$ = 1(12) + 2(1) + 1(16) = 30 g/eq

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

$n=\frac{90.09}{30}=3$

Molecular formula = $(CH_2O)_n=(CH_2O)_3=C_{3}H_{6}O_3$

Thus, the molecular formula of the compound will be, $C_{3}H_{6}O_3$

7 0

4 years ago

You have a radioactive isotope with a mass of 400 grams. The sample has a half life of 5 days. How much is left after 25 days ha

Alexeev081 [22]

Answer:

Amount left after 25 days = 12.5 g

Explanation:

Given data:

Mass of sample = 400 g

Half life of sample = 5 days

Mass left after 25 days = ?

Solution:

First of all we will calculate the number of half lives passes in given time period.

Number of half lives = Time elapsed / Half life

Number of half lives = 25 days/ 5 days

Number of half lives = 5

At time zero = 400 g

At 1st half life = 400 g/2 = 200 g

At 2nd half life = 200 g/2 = 100 g

At 3rd half life = 100 g/2 = 50 g

At 4th half life = 50 g/2 = 25 g

At 5th half life = 25 g/2 = 12.5 g

4 0

3 years ago

Now that you have exercised, walked your

ASHA 777 [7]

Answer: The moles of melatonin are 0.000043

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

Given mass of melatonin = 0.01 g

Molar mass of melatonin $(C_{13}H_{16}N_2O_2)$ = 232.28 g/mol

$\text{Number of moles of melatonin}=\frac{0.01 g}{232.28g/mol}=0.000043mol$

Thus the moles of melatonin in 0.01 g are 0.000043

4 0

3 years ago

Acetone has a boiling point of 56.5 celcius. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57

vodomira [7]

Answer:

0.515 g

Explanation:

<em>Acetone (C₃H₆O) has a boiling point of 56.5 °C. How many grams of the acetone vapor would occupy the 250 mL Erlenmeyer flask at 57 °C and 730 mmHg?</em>

<em />

Step 1: Given data

Temperature (T): 57°C

Pressure (P): 730 mmHg

Volume (V): 250 mL

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15 = 57°C + 273.15 = 330 K

Step 3: Convert "P" to atm

We will use the conversion factor 1 atm = 760 mmHg.

730 mmHg × (1 atm/760 mmHg) = 0.961 atm

Step 4: Convert "V" to L

We will use the conversion factor 1 L = 1,000 mL.

250 mL × (1 L/1,000 mL) = 0.250 L

Step 5: Calculate the moles (n) of acetone

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.961 atm × 0.250 L/(0.0821 atm.L/mol.K) × 330 K

n = 8.87 × 10⁻³ mol

Step 6: Calculate the mass corresponding to 8.87 × 10⁻³ moles of acetone

The molar mass of acetone is 58.08 g/mol.

8.87 × 10⁻³ mol × 58.08 g/mol = 0.515 g

8 0

3 years ago