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Fudgin [204]
3 years ago
15

38 An aluminum sample has a mass of 80.01 g and a density of 2.70 g/cm3. According to the data, to what number of significant fi

gures should the calculated volume of the aluminum sample be expressed?(1) 1 (3) 3
(2) 2 (4) 4
Chemistry
2 answers:
Rufina [12.5K]3 years ago
4 0

Answer:

C. 3

Explanation:

To determine the number of significant figures use this memory aid:  If the decimal is Absent count from the Atlantic (right side).  If the decimal is Present count from the Pacific (left side).   Start counting at the first non-zero digit.

The result of a calculation can only contain the lowest number of significant figures given in the problem.  2.70 has 3 significant figures.

Furkat [3]3 years ago
3 0
The answer is (3) 3. The calculation rule of significant figures is when multiplication and division, the significant figures of result is equal to the least of the numbers involved. The significant figure of 2.70 and 80.01 is 3 and 4.
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Which formula equation represents the burning of sulfur to produce sulfur dioxide?
Dmitriy789 [7]

Answer:

option A = S(s) + O₂(g)   →   SO₂ (s)

Explanation:

Chemical equation:

S(s) + O₂(g)   →   SO₂ (s)

when sulfur burned in the presence of oxygen it produce sulfur dioxide. The sulfur dioxide can further react with oxygen to produce sulfur trioxide and then react with water to form sulfuric acid.

Uses of sulfur dioxde:

It is used as a solvent and reagent in laboratory.

Sulfur dioxide is used to produce sulfuric acid.

It is used as a disinfectant

It is also used as a reducing agent.

It is used to preserve the dry food.

7 0
3 years ago
Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given
tekilochka [14]

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

K_a= 1.34*10^{-5

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝)  \alpha = \sqrt{\frac{K_a}{C} }

\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }

\alpha = \sqrt{4.9629*10^{-5}}

\alpha = 7.044*10^{-3

percentage% (∝) = 7.044*10^{-3}*100

= 0.704%

For Part B; where Concentration of B = 7.84*10^{-2 M

\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }

\alpha = \sqrt{1.709*10^{-4} }

\alpha = 0.0130\\

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= 1.92*10^{-2} M

\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }

\alpha = \sqrt{6.802*10^{-4}}

\alpha =0.02608

percentage% (∝) = 0.02608  × 100%

= 2.60%

7 0
3 years ago
Consider the following reaction:
NeX [460]

For the given reaction, according to the Law of Conservation of Energy, the energy required to decompose Hcl and produce H_{2}+c l_{2} are equal.

Answer: Option C

<u>Explanation:</u>

According to law of conservation's of energy, energy can only be transferred from reactants to product side. So in this process, it is stated that 185 kJ of energy will be needed to decompose it. So that 185 kJ of energy will be getting transferred to produce the creation of hydrogen and chloride in the product side.

So if we see from the reactants side, the energy of 185 kJ is required for decomposition of hydrogen chloride. Similarly, if we see from the product side, the 185 kJ utilized for decomposition is transferred as energy required to create hydrogen and chlorine atoms. This statement will be in accordance with the law of conservation's of energy.

3 0
3 years ago
Read 2 more answers
Which contribution added to Rutherford's discovery of a positive nucleus?
irina [24]
Your answer is: C. Neutrons are inside the nucleus of a atom
6 0
3 years ago
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Controls are defined as ____​
wariber [46]

Answer:

The second one is the answer

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