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3 years ago

Please help me it due to at 6:00pm will mark the brainiest and more points please help me really need help please

Chemistry

2 answers:

emmasim [6.3K]3 years ago

6 0

Answer:

The 3rd Law

Explanation:

The rockets burns fuel, pushing down onto the ground which then creates equal force pushing it up.

12345 [234]3 years ago

4 0

Answer:

the answer is newtons 3rd law of motion (number 3)

Explanation:

You might be interested in

#15 fill in the following table regarding ions

Natasha_Volkova [10]

Answer:

$P^{3-}$ Anion Gained 3 18

$Sr^{2+}$ Cation Lost 2 36

$Fe^{3+}$ Cation Lost 3 23

Explanation:

Looking at the ionic notation, the negative symbol (-) indicates that there are more electrons than protons and the positive symbol (+) indicates that there are more protons than electrons.

The number tells you how many was gained or lost.

To determine how many electrons are left, you base this off how many protons there are. The number of protons in an atom is expressed by the atomic number. In a stable atom, you have an equal number of protons and electrons.

Ions occur when electrons are gained or lost.

A cation is positively charged because it LOST an electron. Since there are more protons than electrons, the charge would be positive.

An anion is negatively charged because it gained an electron. Since there are more electrons than protons, the charge would be negative.

Using the explanation above, you can see how the answers were obtained. As for the last column, just use basic math to do this.

The atomic number of Phosphorus (P) is 15, so this means that there are 15 protons. Since it gained 3 electrons, just add 3 to 15:

15 + 3 = 18

Sr has an atomic number 38. Since it lost 2 electrons, just subtract 2 from the atomic number.

38-2 = 36

Fe has an atomic number 26. It lost 3 electrons, so we subtract again.

26-3 = 23

7 0

3 years ago

Decide which of the following statements are True and which are False about equilibrium systems:A large value of K means the equ

ivanzaharov [21]

Answer:

a. True

b. False

c. True

d. False

e. False

Explanation:

A. (true) The equilibrium constant K is defined as

$\frac{Products}{reagents}$

In any case

aA +Bb ⇌ Cd +dD

where K is:

$K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$

A large value on K means that the concentration of products is bigger than the concentrations of reagents, so the forward reaction is favored, and the equilibrium lies to the right.

B. (False) When we work with gases, we use partial pressure to make calculations in the equilibrium, so we estimate Kp as:

$Kp= \frac{(P_{C})^{c}(P_{D})^{d}}{(P_{A})^{a}(P_{B})^{b}}$

Using the ideal gas law, we can get a relationship between K and Kp

Pv=nRT where $P=\frac{n}{v}*RT$ we know that $\frac{n}{v}$ is the molar concentration. When we replace P in the expression for Kp we get:

$Kp= \frac{[C]^{c}*(RT)^{c}[D]^{d}*(RT)^{d}}{[A]^{a}*(RT)^{a}[B]^{b}*(RT)^{b}}$

Reorganizing the equation:

$Kp= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}*\frac{(RT)^{c+d}}{(RT)^{a+b}}$

We can see K in the expression

$Kp= K*(RT)^{c+d-a-b}$

Delta n = c+d-a-b

$Kp= K*(RT)^{delta n}$

For the reaction

$H_{2}(g) + F_{2}(g)-- equilibrium---2HF(g)$

Delta n = 2-1-1=0

$Kp= K*(RT)^{0}$

So Kp=K in this case.

C. (true) The value of K just depends on the temperature that’s why changing the among of products won’t have any effect on its value.

D. (false) as we can see this reaction involve a heterogeneous system with solids and gases. For convention the concentration for solids and liquids can be considered constant during the reaction that’s why they’re not include in the calculation for the equilibrium constant. Taking this into account the expression for the equilibrium for this reaction is:

$CaCO_{3}(s)---equilibrium----CaO(s) + CO_{2}(g)$

$K= [CO_{2}]$

So we can see that $[CaCO_{3}]$ is not include in the expression.

E. (False) The equilibrium is defined as the point where the rate of the forward reaction is the same to the rate of the reverse reaction. The value of K is telling you which reaction is favored but the rate of both reactions is the same in this point. (see picture)

3 0

3 years ago

The table shows several useful salts formed from reactions between different acids and bases.

ollegr [7]

The two chemical formulas which will correctly complete the table for lithium fluoride are: A. HF, LiOH.

<h3>What is a chemical reaction?</h3>

A chemical reaction can be defined as a chemical process that involves the continuous transformation (rearrangement) of the ionic, atomic or molecular structure of a chemical element by breaking down and forming chemical bonds, in order to produce a new chemical compound while new bonds are formed.

This ultimately implies that, chemical bonds between atoms of a chemical element are broken and then new bonds are formed in all chemical reactions.

<h3>What is a neutralization reaction?</h3>

A neutralization reaction can be defined as a type of chemical reaction between an acid and a base, which typically leads to the formation of an ionic compound (salt) and water as end products as shown below:

LiOH(aq) + HF(aq) → LiF(s) + H₂O(l)

In this context, we can infer and logically deduce that the two chemical formulas which will correctly complete the table for lithium fluoride are HF and LiOH.

Read more on neutralization reaction here: brainly.com/question/27745033

#SPJ1

3 0

2 years ago

Determine the oxidation state for each of the elements below. The oxidation state of ... silver ... in ... silver oxide Ag2O ...

KatRina [158]

Answer:

The oxidation state of silver in $\rm Ag_2O$ is $+1$ .

The oxidation state of sulfur in $\rm SO_2$ is $+4$ .

Explanation:

The oxidation states of atoms in a compound should add up to zero.

There are two silver $\rm Ag$ atoms and one oxygen $\rm O$ atom in one formula unit of $\rm Ag_2O$ . Therefore:

$\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}$ .

The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is $-2$ . Silver oxide $\rm Ag_2O$ isn't an exception. Therefore:

$\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}$ .

Solve this equation for the (average) oxidation state of $\rm Ag$ :

$\text{Oxidation state of $\rm Ag$} = 1$ .

Similarly, because there are one sulfur $\rm S$ atom and two oxygen $\rm O$ atoms in each $\rm SO_2$ molecules:

$\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}$ .

The oxidation state of $\rm O$ in $\rm SO_2$ is also $-2$ , not an exception, either.

Therefore:

$\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}$ .