What is a metamorphic rock?

Answers

Answer 1

Answer: Metamorphic rock is a rock formation derived from the processing of other metamorphic rocks. Due to the high temperatures and pressures that exist in the interior of the Earth, the rocks undergo transformations in its structure and in its mineral composition.

hope this helps!

Related Questions

Olive oil has a density of 0.92 g/mL. How much would 1 liter of olive oil weigh in grams?
Methane reacts with oxygen to produce carbon dioxide and water. What are the products? A.methane and waterB.oxygen and carbon dioxideC.methane and oxygenD.carbon dioxide and wate
The amount of heat transferred from an object depends on which of the following?a. the specific heat of the object b. the initial temperature of the object c. the mass of the object d. all of the above
5. Which of the following is a synthesis reaction? (2 points)2H2 + O2 → 2H2O 2H2O2 → 2H2O + O2 CaCO3 → CaO + CO2 2AgNO3 + Zn → 2Ag + Zn(NO3)2
What is the force that accelerates objects towards Earth?

What is the net ionic equation? (This is already balanced, just need the net ionic) 2CH3COOH(aq)+ Sr(OH)2(aq)----> Sr(CH3COO)2(aq)+2H2O(l)

Answers

Net ionic equation should be like this :
CH3COOH(aq) + OH- --> CH3COO- + H2O(l)
According to the fact that acetic acid is too weal and non-ionized, put CH3COOH, Sr(OH)2 is not so strong but can dissolve therefore can ionize.

Bombarding sodium-23 with a
proton produces nuclide X and a neutron. What is nuclide X?

Answers:

magnesium-24

magnesium-23

neon-23

s
odium-24

none of the above

The isotope P
has a half-life of 14.3 days. If a sample originally contained 1.00 g of P,
how much was left after 43 days?

Answers:

0.250 g

0.125 g

0.750 g

0.500 g

Identify X in the reaction
below.
U
+ C
→ Cf
+ X

Answers:

1
alpha particle

3
protons

6
neutrons

6
electrons

A .20 gram sample of C-14 was allowed to decay for 3 half-lives. What mass
of the sample will remain? Carbon-14 has a half life of 5730
years.

Answers:

0.025

0.05

0.1

0
.01

0.05

The isotope Cu
has a half-life of 30 s. If a sample originally contained 48 mg of Cu,
how much time passed before the amount fell to 3 mg?

Answers:

120 s

240s

30 s

60 s(when I did my calculation for the question above, I got 60 seconds)

What radionuclide decays to Fe-56
by beta emission?

Answers:

Fe

Co

Mn

Co

MnHow do I know wat it becomes. I put 57 Co 27 and it's wrong.

The Cf
to Cf
conversion is accompanied by __________.

Answers:

an alpha
emission

a
neutron capture

an electron
capture

an electron releaseI would greatly appreciate your help...ASAP. Thank you.

Answers

Bombarding sodium-23 with a proton produces nuclide X and a neutron. What is nuclide X?

$^(23)_(11)Na--^(\beta^+)--> ^(23)_(10)Ne + e^(+)$

answer: neon
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The isotope P has a half-life of 14.3 days. If a sample originally contained 1.00 g of P,
how much was left after 43 days?

0 days >>> 1.00g
14,3 days >>> 1.00g :2 = 0,5g
(14.3+14.3) days >>> 0.5g : 2 = 0.25g
(14.3+14.3+14.3) days >>> 0.25g : 2 = 0.125g

answer: 0.125g
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Identify X in the reaction
below.
U + C → Cf + X

$^(238)_(92)U ---> ^(251)_(98)Cf+6e^-$

answer: 6 electrons (so it's beta⁻)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

A .20 gram sample of C-14 was allowed to decay for 3 half-lives. What mass
of the sample will remain? Carbon-14 has a half life of 5730
years.

at the beginning >>> 0.20g
1 half-live >>> 0.20g : 2 = 0.10g
2 half-live >>> 0.10g : 2 = 0.05g
3 half-live >>> 0.05g : 2 = 0.025g

answer: 0.025g
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The isotope Cu has a half-life of 30 s. If a sample originally contained 48 mg of Cu,
how much time passed before the amount fell to 3 mg?

0 s (at the beginning) >>> 48mg
30s >>>>>>>>>>>>>>> 48mg : 2 = 24mg
(30+30)s >>>>>>>>>>> 24mg : 2 = 12mg
(30+30+30)s >>>>>>>> 12mg : 2 = 6mg
(30+30+30+30)s >>>>> 6mg : 2 = 3mg

answer: 30+30+30+30 = 120s
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
What radionuclide decays to Fe-56 by beta emission?

$^(56)_(26)Fe--^(\beta^-)-->^(56)_(27)Co + e^-$

answer: ⁵⁶₂₇Co
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The Cf to Cf conversion is accompanied by __________.

Cf to Cf ???? maybe...mistake?

A substance that can accept a pair of electrons to form a covalent bond is known as a lewis ____.a. base
b. buffer
c. acid
d. catalyst

Answers

c. acid
have a nice day

A container is filled with 4.0 g H2 and 5.0 02. How much water is produced

Answers

4.0 g water. Water is the limiting reactant. Therefore there will be excess oxygen.

A sample of 0.3257 g of an ionic compound containing the bromide ion (Br−) is dissolved in water and treated with an excess of AgNO3. If the mass of the AgBr precipitate that forms is 0.7165 g, what is the percent by mass of Br in the original compound?

Answers

Answer:

Explanation:

Given parameters:

Mass of ionic compound = 0.3257g

Mass of AgBr precipitate = 0.7165g

Unknown:

Percent mass of Br in the original compound.

Solution

The percent mass of Br in original compound = $(mass of Br in the sample)/(mass of sample)$

Now we have to find the mass of Br⁻:

We must note that the same mass of Br⁻ would move through the ionic sample to form the precipitate.

Mass of Br in AgBr = $(Atomic mass of Br)/(Molar mass of AgBr) x mass of precipitate$

Mass of Br = $(80)/(80 + 108)$ x 0.7165

Mass of Br = 0.426 x 0.7165 = 0.305g

Percent mass of Br = $(0.305)/(0.3257)$ x 100 = 93.7%

A sample of nitrogen occupies 10.0 liters at 25°C and 98.7 kPa. What would be the volume at 20°C and 102.7 kPa?

Answers

To solve the problem, we assume the sample to be ideal. Then, we use the ideal gas equation which is expressed as PV = nRT. From the first condition of the nitrogen gas sample, we calculate the number of moles.

n = PV / RT
n = (98.7x 10^3 Pa x 0.01 m^3) / (8.314 Pa m^3/ mol K) x 298.15 K
n = 0.40 mol N2

At the second condition, the number of moles stays the same however pressure and temperature was changed. So, the new volume is calculated as follows:

V = nRT / P
V = 0.40 x 8.314 x 293.15 / 102.7 x 10^3
V = 9.49 x 10^-3 m^3 or 9.49 L

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