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RoseWind [281]
3 years ago
13

Which of the following is the correct name for the compound HF? Hydrogen floride hydrogen monofloride Hydroflorous acid Hydroflo

ric acid
Chemistry
1 answer:
Nataliya [291]3 years ago
4 0
Answer: The correct answer is Hydrogen Fluoride.
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What is the name of a solution whose concentration of solute is equal to the maximum concentration that is predicted from the so
faltersainse [42]

<u>Answer:</u> The correct answer is saturated solution.

<u>Explanation:</u>

For the given options:

Dilute solutions are defined as the solutions in which solute particles are present in less very amount than the solvent particles.

Unsaturated solutions are defined as the solutions where more and more of solute particles can be dissolved in the given amount of solvent.

Saturated solutions are defined as the solutions where no more solute particles can be dissolved in the solvent.  The concentration of the solute particles that can be dissolved in a solution is maximum.

Supersaturated solutions are defined as the solutions where more amount of solute particles are present than the solvent particles.

From the above information, we conclude that the given solution is saturated solution.

7 0
3 years ago
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Does the identity of an atom change if we add or subtract electrons or neutrons? Explain.
AURORKA [14]
Adding or removing neutrons from the nucleus are how isotopes are created. Protons carry a positive electrical charge and they alone determine the charge of the nucleus. Adding or removing protons from the nucleus changes the charge of the nucleus and changes that atom's atomic number.
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2 years ago
A student is given the question: "what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume? the density of gold is 19.3
IRINA_888 [86]

<em>The mass of a gold bar  = 1.424 x 10⁴ gram</em>

<em></em>

<h3><em>Further Explanation</em></h3>

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than type

The unit of density can be expressed in g / cm3 or kg / m3

Density formula:

\large{\boxed{{\bold{\rho~=~\frac{m}{V}}}}

ρ = density

m = mass

v = volume

A common example is the water density of 1 gr / cm3

The mass itself is often equated with weight, even though it is different. Mass is the amount of matter in the matter while weight is related to the gravitational force

<em>Known variable</em>

volume gold bar 7.379 × 10⁻⁴ m³

a density of 19.3 g/cm³

<em>Asked</em>

the mass of a gold bar

<em>Answer</em>

volume is known to be 7.379 × 10⁻⁴ m³, then we change it first to units of cm³ adjusting to units of density

7.379 × 10⁻⁴ m³ = 7.379 × 10² cm³

then:

mass = volume x density

mass = 7.379 × 10² cm³ x 19.3 g/cm³

mass = 1.424 x 10⁴ gram

<h3><em>Learn more </em></h3>

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Keywords: density, mass, volume, a gold bar

3 0
3 years ago
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In order to prepare 50.0 mL of 0.100 M NaOH you will add _____ mL of 1.00 M NaOH to _____ mL of water
FinnZ [79.3K]

The question requires us to complete the sentence regarding the preparation of a more dilute NaOH solution (0.100 M, 50.0 mL) from a more concentrated NaOH solution (1.00 M).

Analyzing the blank spaces that we need to fill in the sentence, we can see that we must provide the volume of the more concentrated solution and the volume of water necessary to prepare the solution.

We can use the following equation to calculate the volume of more concentrated solution required:

\begin{gathered} C_1\times V_1=C_2\times V_2 \\ V_1=\frac{C_2\times V_2}{C_1} \end{gathered}

where C1 is the concentration of the initial solution (C1 = 1.00 M), V1 is the volume required of the inital solution (that we'll calculate), C2 is the concentration of the final solution (C2 = 0.100 M) and V2 is the volume of the final solution (V2 = 50.0 mL).

Applying the values given by the question to the equation above, we'll have:

\begin{gathered} V_1=\frac{C_2\times V_2}{C_1} \\ V_1=\frac{0.100M_{}\times50.0mL_{}}{1.00M_{}}=5.00mL \end{gathered}

Thus, we would need 5.00 mL of the more concentrated solution.

Since the volume of the final solution is 50.0 mL and it corresponds to the volume of initial solution + volume of water, we can calculate the volume of water necessary as:

\begin{gathered} \text{final volume = volume of initial solution + volume of water} \\ 50.0mL=5.00mL\text{ + volume of water} \\ \text{volume of water = 45.0 mL} \end{gathered}

Thus, we would need 45.0 mL of water to prepare the solution.

Therefore, we can complete the sentence given as:

<em>"In order to prepare 50.0 mL of 0.100 M NaOH you will add </em>5.00 mL<em> of 1.00 M NaOH to </em>45.0 mL<em> of water"</em>

5 0
1 year ago
What is a metalloid?
Veronika [31]
an element whose properties are intermediate between those of metals and solid nonmetals. they are electrical semiconductors.
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