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Stella [2.4K]
3 years ago
6

Will a crayon dissolve in water?

Chemistry
2 answers:
sweet-ann [11.9K]3 years ago
6 0

Answer:

Yes

Explanation:Like wax crayons but they can dissolve in water.

erma4kov [3.2K]3 years ago
3 0

Answer:

<h2>here you go :)</h2>

Explanation:

Two variations on watercolor pencils available are woodless pencils (just the pencil 'lead' with a paper wrapper) and water-soluble crayons (like wax crayons, but they dissolve in water). Water-soluble crayons enable you to put down more pigment (or color) faster than a watercolor pencil, as they're softer and broader.

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When 4.41g of phosphoric acid (H3PO4) react with 9.25g of barium hydroxide, water and insoluble barium phosphate form. [T/I-7] a
AnnZ [28]

Answer:

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

Explanation:

Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.

H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

We will balance it using the trial and error method.

First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)

Finally, we will get the balanced equation by multiplying H₂O by 6.

2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)

3 0
3 years ago
Why do atoms share electrons in covalent bonds??
ra1l [238]
Atoms in covalent bonds do combine so as to be stable. As covalent bond consist non metals e.g O2 in this example each atom has vacance of 2 orbitals/ electrons so shairing electrons result their stability
6 0
4 years ago
Read 2 more answers
33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b
Darina [25.2K]

Explanation:

Mass of fructose = 33.56 g

Mass of water =  18.88  g

Total mass of the solution =  Mass of fructose + Mass of water = M

M = 33.56 g + 18.88  g =52.44 g

Volume of the solution = V = 40.00 mL

Density =\frac{Mass}{Volume}

a) Density of the solution:

\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol

Molar mass of water = 18.02 g/mol

Moles of water= n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol

Mole fraction of fructose in this solution:\chi_1

\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}

\chi_1=0.1510

Mole fraction of water = \chi_2=1-\chi_1=0.8490

c) Average molar mass of of the solution:

=\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol

=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

\frac{\text{Average molar mass}}{\text{Density of the mass}}

=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol

5 0
3 years ago
RIGHT ANWSER WILL BE MARKED BRAINLIEST
dybincka [34]

Answer:

11460

Explanation:

4 0
3 years ago
After he conducted cathode ray tube experiments proving the existence of negatively charged particles we now call electrons, Tho
Lina20 [59]

Answer:

Answer is explained below;

Explanation:

In 1904, after the discovery of the electron, the English physicist Sir J.J. Thomson proposed the plum pudding model of an atom. In this model, the atom had a positively-charged space with negatively charged electrons embedded inside it i.e., like a pudding (positively charged space) with plums (electrons) inside.

In 1911, another physicist Ernest Rutherford proposed another model known as the Rutherford model or planetary model of the atom that describes the structure of atoms. In this model, the small and dense atom has a positively charged core called the nucleus. Also, he proposed that just like the planets revolving around the Sun, the negatively charged electrons are moving around the nucleus.

By conducting a gold foil experiment, Rutherford disproved Thomson's model. In this experiment, positively charged alpha particles emitted from a radioactive source enclosed within a protective lead were used which was then focused into a narrow beam. It was then passed through a slit in front of which a thin section of gold foil was placed. A fluorescent screen (coated with zinc sulfide) was also placed in front of the slit to detect alpha particles which on striking the fluorescent screen would produce scintillation (a burst of light) which was visible through a microscope attached to the back of the screen.

He observed that most of the alpha particles passed straight through the gold foil without any resistance and this implied that atoms contain a large amount of open space. The slight deflection of some of the alpha particles, the large-angle scattering of other alpha particles and even the bouncing back of a very few alpha particles toward the source suggested their interactions with other positively charged particles inside the atom.

So, he concluded that only a dense and positively charged particle such as the nucleus would be responsible for such strong repulsion. Also, the negatively charged electrons electrically balanced the positive nuclear charge and they moved around the nucleus in circular orbits. Between the electrons and nucleus, there was an electrostatic force of attraction just like the gravitational force of attraction between the sun and the revolving planets.

Later, the Rutherford model was replaced by the Bohr atomic model.

6 0
3 years ago
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