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Montano1993 [528]
3 years ago
6

This is a graded Quiz and I need all of the answers to be right but I am not sure so I need help this is a 3.08 science Quiz and

it's do today. Thank you for the help
Question 1 (1 point)

True or False

Intrusive igneous rocks have fine-grained crystals.

Tue
False


Question 2 (1 point)

True or false

Igneous rocks are formed from cooling and hardening magma or lava.

True
False


Question 3 (1 point)

True or false

Metamorphic rocks are formed when sediments harden it to rocks.

True
False


Question 4 (1 point)

True or false

Lithification is the process of converting sediments into solid rock through compaction and cementation.

True
False


Question 6 (1 point)

What type of rock forms when heat and pressure change existing rocks WITHOUT melting them?


Metamorphic

Sedimentary

Igneous



Question 7 (1 point)

Once a metamorphic rock, always a metamorphic rock. cannot change from one type of rock to another

True
false


Question 8 (1 point)

The rock cycle has a definitive beginning and end

true
false


Question 9 (1 point)


Foliation occurs in sedimentary rocks

True
false


Some of the words are not spelled right and that's all my teachers fault i'm sorry.
Also if you are the first to answer and you answered all the of the questions and answered them right you will get marked brainly whatever it's called just please answer these all true fully and right and not just pick one for the points thank you bye
Chemistry
1 answer:
Setler79 [48]3 years ago
8 0

Answer:

1 true

Explanation:

bcz it is true

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What is 0.00550 g converted to mg
Mila [183]

Answer: it is 5.5 mg

Explanation:

you have to multiply the mass value by 1000

5 0
3 years ago
In the electron cloud model, the electron cloud is denser in some locations than in others.
Sonbull [250]
<span>In the electron cloud model, the denser areas represent that there is a great probability that a good number of electrons are ganged up or crowded in that area. The electrons affect the density of some parts of the electron cloud when they condense in those locations.</span>
8 0
3 years ago
Hydrocarbons do not dissolve in concentrated sulfuric acid, but methyl benzoate does. Explain this difference and write an equat
finlep [7]

Answer:

See explanation

Explanation:

For a substance to dissolve in another, there must be some sort of interaction between the substances.

Recall that like dissolves like. That is, polar substances dissolve polar substances and non polar substances dissolve nonpolar substances.

Hydrocarbons are nonpolar hence they do not dissolve in polar sulphuric acid. Methyl benzoate is polar hence it dissolve in polar sulphuric acid.

The equation showing the ions is depicted in the image attached to this answer.

7 0
3 years ago
How does the A Hreaction relate to the A He of molecules involved in a reaction?
igor_vitrenko [27]

Answer:

B. ΔHreaction = ΔH°f reactants- ΔH°f products

Explanation:

<em>Using Hess's law, it is possible to sum ΔH of several related reactions to find ΔH of a particular reaction</em>.

Having in mind Hess's law, ΔH°f is defined as the change in enthalpy during the formation of 1 mole of substance from its constituent elements (That is, pure elements, mono or diatomics, that have a ΔH° = 0).

For example, in ΔH°f of H₂O, the equation is:

H₂(g) + 1/2O₂(g) → H₂O(g)

The constituent elements with ΔH°f = 0 are H₂(g) and O₂(g).

Now, using Hess's law, you can sum the ΔH°f of substance in a reaction as, for example:

NaOH + HCl → H₂O + NaCl. ΔHr

The ΔH°f for each substance in the reaction is:

NaOH: Na + 1/2H₂ + 1/2O₂ → NaOH <em>(1)</em>

HCl: 1/2H₂ + 1/2Cl₂ → HCl <em>(2)</em>

H₂O: H₂ + 1/2O₂ → H₂O <em>(3)</em>

NaCl: Na + 1/2Cl₂ → NaCl <em>(4)</em>

The algebraic sum of (3) + (4) is -(ΔH°f reactants):

H₂ + 1/2O₂ + Na + 1/2Cl₂ → NaCl + H₂O ΔH°f reactants

This reaction - {(1)+(2)} ΔH°f products

NaOH + HCl → H₂O + NaCl.

ΔHr = ΔH°f reactants- ΔH°f products

In the example, we obtain this relationship that can be expanded for all reactions. Thus, right answer is:

<h3>B. ΔHreaction = ΔH°f reactants- ΔH°f products</h3>

8 0
3 years ago
Consider the reaction: N2(g) + O2(g) ⇄ 2NO(g) Kc = 0.10 at 2000oC Starting with initial concentrations of 0.040 mol/L of N2 and
IrinaVladis [17]

Answer:

0.011 mol/L

Explanation:

This can be solved with something called an ICE table.

I = initial

C = change

E = equilibrium

Initially, there is 0.04 M of N₂, 0.04 M of O₂, and 0 M of NO.

x amount of N₂ reacts.  Since the stoichiometry is 1:1, x amount of O₂ also reacts.  This produces 2x of NO.

After the reaction, there is 0.04-x of N₂, 0.04-x of O₂, and 2x of NO.

Here it is in table form:

\left[\begin{array}{cccc}&N2&O2&NO\\I&0.04&0.04&0\\C&-x&-x&+2x\\E&0.04-x&0.04-x&2x\end{array}\right]

Now we can use the equilibrium constant:

Kc = [NO]² / ( [N₂] [O₂] )

Substituting:

0.10 = (2x)² / ( (0.04 - x) (0.04 - x) )

Solving:

0.10 = (2x)² / (0.04 - x)²

√0.10 = 2x / (0.04 - x)

(√0.10) (0.04 - x) = 2x

(√0.10)(0.04) - (√0.10)x = 2x

(√0.10)(0.04) = 2x + (√0.10)x

(√0.10)(0.04) = (2 + √0.10)x

x = (√0.10)(0.04) / (2 + √0.10)

x = 0.0055

At equilibrium, the concentration of NO is 2x.  So the answer is:

[NO] = 2x

[NO] = 0.011

The equilibrium concentration of NO is 0.011 mol/L.

3 0
3 years ago
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