All categories

3 years ago

What is the most likely metal ion present in the fireworks shown in the image?

Chemistry

2 answers:

Tasya [4]3 years ago

8 0

A because of the way it looks and how it works

Inga [223]3 years ago

7 0

Answer:

It's A I just took the test

Explanation:

You might be interested in

Elaborate on how to classify elements and pure compounds.

tankabanditka [31]

Answer: C) Elements and pure compounds are homogeneous materials because they have a uniform composition throughout.

Explanation: Element is a pure substance which is composed of atoms of similar elements. Compound is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass.

Elements are compounds form homogeneous materials as they have uniform composition throughout and the components are evenly distributed throughout the material.

Mixtures are heterogeneous materials as they do not have uniform composition and the components are not evenly distributed throughout the material.

4 0

3 years ago

Read 2 more answers

Nitric acid can be formed in two steps from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming natur

worty [1.4K]

Answer:

-376 kJ.

Explanation:

4 0

3 years ago

The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate

Delicious77 [7]

Answer:

For a: The edge length of the crystal is 533.5 pm

For b: The atomic radius of potassium is 231.01 pm

Explanation:

For a:

To calculate the lattice parameter or edge length of the crystal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $0.855g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal = 39.09 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell = ?

Putting values in above equation, we get:

$0.855=\frac{2\times 39.09}{6.022\times 10^{23}\times (a)^3}\\\\\a=5.335\times 10^{-8}cm=533.5pm$

Conversion factor: $1cm=10^{10}pm$

Hence, the edge length of the crystal is 533.5 pm

For b:

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 533.5 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 533.5}{4}=231.01pm$

Hence, the atomic radius of potassium is 231.01 pm

4 0

3 years ago

An experiment was conducted to determine the density of a rock sample. The table shows a partial record of the experiment.

mart [117]

Answer: the density of the rock

55.91g/ml

Explanation:

The density of the rock after it had being placed in the cylinder of water so the calculation should look like this:

Volume of water substract the mass of the rock:

And that is 142.5 ml - 86.59g =

Answer 55.91 g/ml

So the density of the rock is 55.91g/ml

5 0

3 years ago

Benzene is a minor component of gasoline. The standard molar enthalpy of formation of benzene C7H16(l) is 48.95 kJ/mol. For the

Radda [10]

Answer:

-3135.47 kJ/mol

Explanation:

Step 1: Write the balanced equation

C₆H₆(l) + 7.5 O₂(g) ⇒ 6 CO₂(g) + 3 H₂O(g)

Step 2: Calculate the standard enthalpy change of the reaction (ΔH°r)

We will use the following expression.

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpies of formation

p: products

r: reactants

ΔH°r = 6 mol × ΔH°f(CO₂(g)) + 3 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₆H₆(l)) - 7.5 mol × ΔH°f(O₂(g))

ΔH°r = 6 mol × (-393.51 kJ/mol) + 3 mol × (-241.82 kJ/mol) - 1 mol × (48.95 kJ/mol) - 7.5 mol × 0 kJ/mol

ΔH°r = -3135.47 kJ

Since this enthalpy change is for 1 mole of C₆H₆(l), we can express it as -3135.47 kJ/mol.

7 0

3 years ago