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3 years ago

An atom that gains or loses an electron has a net electric charge and is called a/n

Chemistry

2 answers:

Fofino [41]3 years ago

8 0

That would be an ion, so A.

uysha [10]3 years ago

7 0

I believe the answer is A

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Which of the following is NOT a factor that determines how much energy we need?

german

Answer:

B) The amount of vitamin D gained in 1 day.

Explanation:

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2 years ago

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How much pure water must mixed with 8 pints of 50 developer to produce a mixture that is 24%?

Galina-37 [17]

8:24 = 0.333(3) pints is one percent
0.333(3)* 100=33.333(3) pints will be 24% mixture with the water
33.3333-8=25.333 pints of water is required for producing 24% mixture

25.3333 pints of pure water and 8 pints of juce.

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3 years ago

What is the mass in grams of ba(io3)2 can be dissolved in 500 ml of water at 25 degrees celcius?

lesya [120]

The mass of Ba(IO3)2 that can be dissolved in 500 ml of water at 25 degrees celcius is 2.82 g

<h3>What mass of Ba(IO3)2 can be dissolved in 500 ml of water at 25 degrees celcius?</h3>

The Ksp of Ba(IO3)2 = 1.57 × 10^-9

Molar mass of Ba(IO3)2 = 487 g/mol?

Dissociation of Ba(IO3)2 produces 3 moles of ions as follows:

$Ba(IO_{3})_{2} \leftrightharpoons Ba^{2+} + 2\:IO_{3}^{-}$

$Ksp = [Ba^{2+}]*[IO_{3}^{-}]^{2}$

$[Ba(IO_{3})_{2}] = \sqrt[3]{ksp} =\sqrt[3]{1.57 \times {10}^{ - 9} } \\ [Ba(IO_{3})_{2}] = 1.16 \times {10}^{-3} moldm^{-3}$

moles of Ba(IO3)2 = 1.16 × 10^-3 × 0.5 = 0.58 × 10^-3 moles

mass of Ba(IO3)2 = 0.58 × 10^-3 moles × 487 = 2.82 g

Therefore, mass Ba(IO3)2 that can be dissolved in 500 ml of water at 25 degrees celcius is 2.82 g.

Learn more about mass and moles at: brainly.com/question/15374113

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2 years ago

What is the most important reason to consider ethics when conducting

Anon25 [30]

Answer:

B. It is important that people are not harmed for the sake of science.

Explanation:

Ethical principles stress the need to do good and cause no harm.A researcher is therefore required to;

obtain an informed consent from the participants
minimize or eliminate risk of harm to participants
protect the anonymity and confidentiality of participants
Apply no deceptive techniques
allow the right to withdraw from the study by a participant

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3 years ago

During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce

swat32

Answer: 16.32 g of $O_2$ as excess reagent are left.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol$

$\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol$

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

According to stoichiometry :

2 moles of $SO_2$ require = 1 mole of $O_2$

Thus 0.34 moles of $SO_2$ will require= $\frac{1}{2}\times 0.34=0.17moles$ of $O_2$

Thus $SO_2$ is the limiting reagent as it limits the formation of product and $O_2$ is the excess reagent.

Moles of $O_2$ left = (0.68-0.17) mol = 0.51 mol

Mass of $O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g$

Thus 16.32 g of $O_2$ as excess reagent are left.

3 0

3 years ago