How many mL of a 5.0% glucose solution provide 80.0 g of glucose?

Answer:

c. 6.

Explanation:

Based on the question's description, the elements must be from the p-block of the periodic table and in the fifth period. They should have valence electrons located in the 5p orbital.

The following elements from the p-block of period 5 meet these criteria: Sr, In, Sn, Sb, Te, and I.

Therefore, there are a total of six such elements.

Answer:

The mass percentage of GaBr₃ in the solid mixture is 30.2 %.

Explanation:

GaBr₃(aq) + 3 AgNO₃(aq) ⟶ 3 AgBr(s) + Ga(NO₃)₃(aq)

To find the molar mass, we have: MW GaBr₃ = 309.4 g/mol and MW AgBr = 187.8 g/mol.

Using these values, we note that 187.8 g of AgBr corresponds to 1 mol.

If 0.368 g of AgBr yields x moles, we calculate x to be 2.0 x 10⁻³ mol AgBr.

Based on the stoichiometry, 1 mol of GaBr₃ produces 3 mol of AgBr, giving us:

y ___________ 2.0 x 10⁻³ mol AgBr

Thus, y equals 6.7 x 10⁻⁴ mol GaBr₃.

Additionally, 1 mol of GaBr₃ has a mass of 309.4 g.

Consequently, 6.7 x 10⁻⁴ mol of GaBr₃ corresponds to w grams of GaBr₃, yielding w = 0.206 g GaBr₃.

Relating this to the total mass:

0.6813 g _____ 100%

0.206 g _____ z

Therefore, z is calculated as 30.2 %.

Answer:

1.44 g/mL

Explanation:

The following information was derived from the provided question:

Volume (V) of H2O2 = 50 mL.

Total mass of flask and H2O2 = 88.5 g.

Mass of the flask = 16.5 g.

What is the density (D)?

Next, we will calculate the mass of hydrogen peroxide (H2O2).

This is determined as follows:

Total mass of flask and H2O2 = 88.5 g.

Mass of the flask = 16.5 g.

What is the mass of H2O2?

Mass of H2O2 = (Total mass of flask + H2O2) – (Mass of flask)

Mass of H2O2 = 88.5 – 16.5

Mass of H2O2 = 72 g

Finally, we will find the density of hydrogen peroxide (H2O2) as follows:

Volume (V) of H2O2 = 50 mL.

Mass (m) of H2O2 = 72 g.

What is the density (D)?

Density (D) = mass (m) divided by volume (V)

D = m/V

D = 72 g / 50 mL

D = 1.44 g/mL

In conclusion, the hydrogen peroxide (H2O2) density is 1.44 g/mL.

Answer:

Given that camphor is a sublime material while sand is not, a gradual heating of the mixture allows for the separation of camphor from sand. The camphor vapors can then be collected and allowed to cool. This process will result in the formation of solid camphor crystals.

Explanation:

To accomplish this, we must first understand the rules of significant figures.
<span>Rule #1: All digits other than zero are considered significant. (1234)
 Rule #2: Leading zeros do not count as significant. (0.093)
 Rule #3: Zeros situated between non-zero digits are significant. (78309)
 Rule #4: Trailing zeros are significant only if there is a decimal point. (0.05470)

Therefore, considering these rules, 56.0g contains three significant figures due to the decimal point.
0.0004m has just one significant figure per Rule #2.
1003ml contains 4 significant figures because the zeros are between two significant digits.
Lastly, 0.0350s has 3 significant figures because digits following a decimal point are counted.

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