All categories

3 years ago

Limescale removers contain sulfamic acid, H3NSO3, which reacts with minerals in hard water

Chemistry

1 answer:

I am Lyosha [343]3 years ago

4 0

Answer:

0.375 moles of CaCO₃ are required

Explanation:

Given data:

Number of moles of sulfamic acid = 0.75 mol

Number of moles of calcium carbonate required = ?

Solution:

Chemical equation:

2H₃NSO₃ + CaCO₃ → Ca(SO₃NH₂)₂ + CO₂ + H₂O

Now we will compare the moles of H₃NSO₃ and CaCO₃ .

H₃NSO₃ : CaCO₃

2 : 1

0.75 : 1/2×0.75 = 0.375 mol

Thus, 0.375 moles of CaCO₃ are required.

You might be interested in

You are counting red blood cells, which objective will be best to count the number of cells, high power or low power objective a

BARSIC [14]

Answer:

Low power

Explanation:

Low power would allow for the full image of the red blood cells and would appear as small circles.

4 0

3 years ago

N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?

just olya [345]

Answer:

$\large \boxed{\text{197.4 g}}$

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 28.01 17.03

N₂ + 3H₂ ⟶ 2NH₃

m/g: 240.0

(a) Moles of NH₃

$\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}$

(b) Moles of N₂

$\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}$

(c) Mass of N₂

$\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}$

7 0

3 years ago

Read 2 more answers

At 25.0°c, a solution has a concentration of 3.179 m and a density of 1.260 g/ml. the density of the solution at 50.0°c is 1.249

oksano4ka [1.4K]

Answer: -

3.151 M

Explanation: -

Let the volume of the solution be 1000 mL.

At 25.0 °C, Density = 1.260 g/ mL

Mass of the solution = Density x volume

= 1.260 g / mL x 1000 mL

= 1260 g

At 25.0 °C, the molarity = 3.179 M

Number of moles present per 1000 mL = 3.179 mol

Strength of the solution in g / mol

= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)

Now at 50.0 °C

The density is 1.249 g/ mL

Mass of the solution = density x volume = 1.249 g / mL x 1000 mL

= 1249 g.

Number of moles present in 1249 g = Mass of the solution / Strength in g /mol

= $\frac{1249 g}{396.35 g/mol}$

= 3.151 moles.

So 3.151 moles is present in 1000 mL at 50.0 °C

Molarity at 50.0 °C = 3.151 M

7 0

3 years ago

Based on the kinetic molecular theory, which of the following statements is correct about the particles in a sample of gas at a

fgiga [73]

Answer:

There is a lot of empty space between them.

Explanation:

The kinetic molecular theory postulates that a substance is made up of tiny particles called molecules. The molecules of a gas are in constant random motion and collide elastically with each other. They also collide with the walls of the container.

The magnitude of intermolecular forces of attraction between gas molecules is very small. Hence gas molecules are largely apart with a lot of empty space between gas molecules.

8 0

3 years ago

The mass of 2.50 moles of calcium fluoride is ____ grams

Vesna [10]

Answer:

195.187016

Explanation:

3 0

3 years ago

Read 2 more answers