All categories

2 years ago

When a solid dissolves in a liquid, it is termed

1 answer:

5 0

A solid dissolves in a liquid when it mixes completely with the liquid. ... Things which dissolve are called solutes and the liquid in which they dissolve is called a solvent to form a solution

You might be interested in

If 14 moles of oxygen react with 14 moles of hydrogen to produce water, what is the

Leokris [45]

Hydrogen is the limiting reactant because when doing a stoichiometry equation for the reactants, hydrogen will be used completely by having a smaller yield and oxygen will be excess (7 moles to be exact)

7 0

2 years ago

LINKING IN<br> TECHNICAL OBJECTS<br> 1 a) What is linking?

d1i1m1o1n [39]

Answer:

A link is a fastening unit that attaches two parts of an object together

Different types of links have different characteristics

4 0

2 years ago

Read 2 more answers

What is the most likely meaning of concentration in paragraph 7?thinking thinking about about one one thing thing in in a a focu

astraxan [27]

Answer:

amount of a substance found in water.

6 0

3 years ago

Read 2 more answers

If 3.00 mL of 0.0250 M CuSO4 is diluted to 25.0 mL with pure water, what is the molarity of copper(II) sulfate in the diluted so

g100num [7]

Answer:

0.00268 M

Explanation:

To find the new molarity, you need to (1) find the moles of CuSO₄ (via the molarity equation using the beginning molarity and volume) and then (2) find the new molarity (using the moles and combined volume). Your final answer should have 3 sig figs to match the given values.

3.00 mL / 1,000 = 0.00300 L

Molarity = moles / volume (L)

0.0250 M = moles / 0.00300 L

(0.0250 M) x (0.00300 L) = moles

7.50 x 10⁻⁵ = moles

25.0 mL / 1,000 = 0.0250 L

0.0250 L + 0.00300 L = 0.0280 L

Molarity = moles / volume (L)

Molarity = (7.50 x 10⁻⁵ moles) / (0.0280 L)

Molarity = 0.00268 M

8 0

1 year ago

If 100.0g of nitrogen is reacted with 100.0g of hydrogen, what is the excess reactant? What is the limiting reactant? Show your

Drupady [299]

N₂ : limiting reactant

H₂ : excess reactant

<h3>Further explanation</h3>

Given

mass of N₂ = 100 g

mass of H₂ = 100 g

Required

Limiting reactant

Excess reactant

Solution

Reaction

mol N₂(MW=28 g/mol) :

$\tt mol=\dfrac{mass}{MW}=\dfrac{100}{28}=3.571$

mol H₂(MW= 2 g/mol) :

$\tt mol=\dfrac{100}{2}=50$

A method that can be used to find limiting reactants is to divide the number of moles of known substances by their respective coefficients, and small or exhausted reactans become a limiting reactants

From the equation, mol ratio N₂ : H₂ = 1 : 3, so :

$\tt \dfrac{3.571}{1}\div \dfrac{50}{3}=3.571\div 16.6$

N₂ becomes a limiting reactant (smaller ratio) and H₂ is the excess reactant

5 0

2 years ago