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3 years ago

What is true about the position of any metal in the periodic table

Chemistry

1 answer:

Serjik [45]3 years ago

7 0

Answer:

The metals are on the bottom left in the periodic table, and the nonmetals are at the top right. The semimetals lie along a diagonal line separating the metals and nonmetals. The elements are arranged in a periodic table, which is probably the single most important learning aid in chemistry.

Explanation:

im not sure if there was supposed to options to that question or not

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If antique brass is 30.0% zinc and 70.0% copper, how many grams of antique brass can be made from 18.0 grams of zinc and 32.8 gr

wariber [46]

Answer: 51.86

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3 years ago

Which energy source contributes to the greatest emissions of gases in the environment during the energy production process

olga_2 [115]

Answer:

Natural gas, emitting fewer harmful chemicals into the atmosphere than other fossil fuels, can help to mitigate some of these environmental issues. These issues include: Greenhouse Gas Emissions. Smog, Air Quality and Acid Rain

Explanation:

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3 years ago

What is the identity of the atom shown?

posledela

Answer:

Explanation:

Argon has 17 protons and 18 neutron

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3 years ago

Read 2 more answers

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

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3 years ago

Based on the results you observed for the iodine test and Benedict’s test, is it better to detect enzyme activity by measuring t

dlinn [17]

Answer:

Is better use the Benedict's test by the increase in the amount of the products if the enzyme is a reductase

Explanation:

The Benedict's test works by the reaction of the reducing sugars with the ion cupric of the reactive. If the enzyme is a reductase (degrades polysaccharides into bi o monosaccharides), it should cut the polysaccharide bond and the products would react with the Benedict's cupric ion

I hope you undestand me

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3 years ago