Which of the following is the correct name for the compound HF? Hydrogen floride hydrogen monofloride Hydroflorous acid Hydroflo
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<em>The mass of a gold bar = 1.424 x 10⁴ gram</em>
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<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:
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
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Keywords: density, mass, volume, a gold bar
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:
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:
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:
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>