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3 years ago

An experiment looking at structures smaller than a cell would most likely employ a

Chemistry

1 answer:

Veseljchak [2.6K]3 years ago

4 0

Answer:

In a conventional optical microscope, objects less than about 200 nanometers apart cannot be distinguished from one another. ... Although electron microscopes produce a detailed image of very small structures, they cannot provide an image of the proteins that make up those structures.

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What is the formula of hexaaquamanganese(ii) sulfate? express your answer as a chemical formula?

Digiron [165]

Answer : The formula of hexaaquamanganese(II)sulfate is $[Mn(H_2O)_6]SO_4$ .

Explanation :

Rules for writing formulas of coordination complexes :

Central metal is written first and then the ligands are written along with their prefixes.

The anion ligands are written before the neutral ligands.

If there are more than one anion ligands or neutral ligands then they are written in alphabetical order.

The given name of complex is, hexaaquamanganese(ii) sulfate.

In this, the central metal is manganese which is written as Mn and the ligand is water (which is a neutral ligand) is written as $H_2O$ and an anion is sulfate which is written as $SO_4$ .

Water is a neutral ligand and sulphate has a charge of -2. As the overall complex is neutral the metal bears a +2 charge which is represented in Roman numeral after the metal.

Therefore, the formula of hexaaquamanganese(II)sulfate is written as $[Mn(H_2O)_6]SO_4$ .

3 0

3 years ago

Balance the equation. Help please!!

Artemon [7]

<h2>Answer:</h2>

Unbalanced equation:

SiO2 + CaC2 → Si + CaO + CO2

Balanced equation:

5SiO2 + 2CaC2 → 5Si + 2CaO + 4CO2

<h3>Explanation:</h3>

The chemical equation needs to be balanced so that it follows the law of conservation of mass.
A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

3 0

4 years ago

Read 2 more answers

As NH4Cl dissolves in water, and cools the solution,

azamat

Answer:

d. delta H is positive, and delta S is positive

Explanation:

The process of dissolution of NH4Cl is:

NH4Cl(s) → NH4+(aq) + Cl-(aq)

As the solution is cooled, the process is absorbing energy, that means ΔH > 0. (ΔH < 0 when the heat is released).

As there are produced 2 ions from 1 solid molecule, the disorder is increasing and ΔS > 0

The right answer is:

<h3>d. delta H is positive, and delta S is positive</h3>

5 0

3 years ago

How many atoms are in 25.00 g of B?

klio [65]

Answer:

There are 1.393 x 10²⁴ atoms in 25.00 g of B.

Explanation:

Hey there!

We are given a value, in grams, that we need to convert to a number of atoms.

We can convert grams to atoms by using Avogadro's Number ( $N_A$ ). This number is equivalent to $6.022 \times 10^{23}$ .

This number can be used to convert any values to:

atoms
molecules
formula units
moles

In order to do this problem, we will need to use dimensional analysis (DA). This process allows us to convert from grams to atoms.

We need to set up our ratios in order to work this out. We can use a periodic table to help us through this next part of the problem.

<u>1. Locating the number of moles of B in the sample</u>

We first need to find the amount of moles of boron (B) there are in the sample.

Checking a periodic table, the atomic mass in atomic mass units (amu) is 10.81 amu.

Atomic mass units can easily be converted to grams and these units can be used interchangeably.

Therefore, for each atom of boron, it weighs 10.81 grams to us. This is equivalent to the mass of one mole of boron.

To find the number of moles, we have two possible ratios we can use:

$\displaystyle \frac{1 \ mole \ B}{10.81 \ grams \ B}$

$\displaystyle \frac{10.81 \ grams \ B}{1 \ mole \ B}$

These ratios mean the same thing, but we need to convert our final unit to moles.

We are given a sample in grams, and when dividing our units, we need to keep moles.

Since the first portion of our expression is in grams, we need to have grams in the bottom of our expression.

$\displaystyle 25.00 \ \text{grams B} \ \times \frac{1 \text{mole B}}{10.81 \ \text{grams B}}$

We can now simplify the expression. Our <u>grams B</u> unit will cancel out, so we are therefore left with <u>moles B</u> remaining.

<u>2. Locating the number of atoms in the sample</u>

Now with our equation, we can convert our number of moles that would be solved if we stopped with the above. However, we need to convert to atoms.

We use Avogadro's number and create a ratio with that of moles.

$\displaystyle \frac{6.022 \times 10^{23}\text{atoms}}{1 \text{mole B}}$

$\displaystyle \frac{1 \text{mole B}}{6.022 \times 10^{23} \text{atoms}}$

We need to cancel out our moles and end with atoms, so we must have moles in the denominator. Therefore, we use the first ratio.

Using our previous expression, we multiply by this new ratio and solve the expression.

$\displaystyle 25.00 \ \text{grams B} \ \times \frac{1 \text{mole B}}{10.81 \ \text{grams B}} \ \times \frac{6.022 \times 10^{23}\text{atoms}}{1 \text{mole B}}$