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3 years ago

Which element is reduced in this reaction? cr2o72−+3hno2+5h+→2cr3++3no3−+4h2o enter the chemical symbol of the element?

1 answer:

3 0

A reducing agent loses electrons, so on the left side of the equation N in HNO2 has an oxidation number of +3 and on the right side in NO3^- it has an oxidation number of +5, so it has lost electrons. Thus, the reducing agent would be HNO2.

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calculate the volume occupied by 10g of propane gas, under normal conditions of temperature and pressure

andriy [413]

Answer:

5.5 L

Explanation:

First we convert 10 g of propane gas (C₃H₈) to moles, using its molar mass:

10 g ÷ 44 g/mol = 0.23 mol

Then we use the PV=nRT formula, where:

P = 1 atm & T = 293 K (This are normal conditions of T and P)

n = 0.23 mol

R = 0.082 atm·L·mol⁻¹·K⁻¹

V = ?

1 atm * V = 0.23 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 K

V = 5.5 L

3 0

3 years ago

Is ammonium ion a bronsted base

Leokris [45]

Unlikely. It's unlikely for ammonium ion ${\text{NH}_4}^{+}$ to accept a proton $\text{H}^{+}$ and act as a Bronsted-Lowry Acid.

<h3>Explanation</h3>

What's the definition of Bronsted-Lowry acids and bases?

Bronsted-Lowry Acid: a species that can donate one or more protons $\text{H}^{+}$ in a reaction.

Bronsted-Lowry Base: a species that can accept one or more protons $\text{H}^{+}$

Ammonium ions ${\text{NH}_4}^{+}$ are positive. Protons $\text{H}^{+}$ are also positive.

Positive charges repel each other, which means that it will be difficult for ${\text{NH}_4}^{+}$ to accept any additional protons. As a result, it's unlikely that ${\text{NH}_4}^{+}$ will accept any proton and act like a Bronsted-Lowry Base.

6 0

3 years ago

The ground state electron configuration of Fe is what

Sindrei [870]

Answer:

Iron?

Explanation:

3 0

2 years ago

snow_tiger [21]

There are 1,000m is 1k. So just move the decimal one position right. 127.56m

There are 10,000cm in 1k. Move the decimal two positions right. 1275.6cm

5 0

3 years ago

A radioisotope decays to give an alpha particle andPb-208.

oee [108]

Answer: The original element is $_{84}^{212}\textrm{Po}$

Explanation:

Alpha decay is defined as the process in which alpha particle is emitted. In this process, a heavier nuclei decays into a lighter nuclei. The alpha particle released carries a charge of +2 units.

The released alpha particle is also known as helium nucleus.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha$

For the given alpha decay process of an isotope:

$^{Z}_{A}\textrm{X}\rightarrow ^{208}_{82}\textrm{Pb}+_2^4\alpha$

To calculate A:

Total mass on reactant side = total mass on product side

A = 208 + 4

A = 212

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

Z = 82 + 2

Z = 84

The isotopic symbol of unknown element is $_{84}^{212}\textrm{Po}$

Hence, the original element is $_{84}^{212}\textrm{Po}$

3 0

3 years ago