Group IA elements are very reactive metal . why ​

Since, the options have not been given the question is incomplete.

What is Darwin's name for species that do not appear to have changed for millions of years?

a.

Dinosaurs

b.

Living fossils

c.

Old souls

d.

Ancient moderns

Answer: b. Living fossil

Explanation:

In 1859 Charles Darwin proposed the term living fossil, that means a species or group of species that had not changed in terms of evolutionary context thus can be useful in tracing the extinct or previously existing forms of life. The examples of the living fossils are horseshoe crabs, ginkgo (Conifers) and tuatara. These group of animals were existed unchanged in the Ordovician, Permian, and Triassic periods respectively with few surviving species.

Answer:

the answer is living fossils so it would be b

Explanation:

how do i know i took the test

invertebrate is a animal that does not have a back bone but may have a exoskeleton examples could be earthworm, jellyfish, snail and a spider because it has a exoskeleton a hard outer shell hope this helps

Vertebrates have a back bone like sea turtles , fish ,foxes and humans.

Answer:

maintaining an internal body temperature is one of the key components to maintaining homeostasis

It's true since homeostasis is the regulation of body temperature

Answer:

0.38

Explanation:

Molar mass of thiophene= 84g/mol

Mass of thiophene = 37g

Number of moles= 37/84= 0.44 moles

Molar mass of heptane= 100 g/mol

Mass of heptane = 72g

Number of moles = 72/100= 0.72 moles

Total number of moles= 0.44 + 0.72= 1.16 moles

mole fraction of thiophene = 0.44/1.16= 0.38

Answer: it would stop producers from producing food.

Explanation: why? Without the action of the producers to turn sunlight into food, the producers would die, and consumers that rely on them would lose their food source and die too.

Answer:

It would stop producers from producing food

Explanation:

Producers would die, and if producers died, consumers would die too.

Hope this helps

Answer:

Molar mass of CaO

(40 + 16)g

= 56g

hope it helps you

To solve this we must be knowing each and every concept related to force. Therefore, the  objects motion will be towards downward with a force of 5 N.

A force is an outside entity that has the power to alter a body's state of rest or motion. It has a direction and a magnitude. The point where forces are applied is determined by the the force's direction as well as the application of the force.

Newton and dyne units are used to measure the force exerted on an item. When using the centimeter kilogram second system of units, force is measured in dynes (CGS unit). In the common international system of units, it is denoted by the letter Newton (N) (SI unit).

upward force=5 N

downward force= 10 N

total force acting on body=downward force-upward force

                                          = 10 N - 5 N

                                          =5 N downward force

The  objects motion will be towards downward with a force of 5 N.

Therefore, the  objects motion will be towards downward with a force of 5 N.

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Answer:

By bonding due to the attractive forces

Answer:

Explanation:

From the combustion of carbon, the reactions occurring in limited oxygen conditions are:

[tex]C(graphite) + \dfrac{1}{2}O_{2(g)} \to CO_{(g)}[/tex]  

[tex]C(graphite) + O_{2(g)} \to CO_{2(g)}[/tex]  

If it occurs in excess, then any leftover CO changes to CO2. i.e.

[tex]C(graphite) + O_{2(g)} \to CO_{2(g)}[/tex]   ---- (1)

[tex]CO_{(g)} + \dfrac{1}{2}O_{(g)} \to CO_{2(g)}[/tex]          ----- (2)

From (1), the enthalpy change is:

[tex]\Delta H_{rxn1} = \Delta H^0_{fCO_2(g)} - ( \Delta H^0_{f C(graphite)}+ \Delta H^0_{fCO_2(g)}[/tex]

[tex]\Delta H_{rxn1} =-393.5 \ kJ/mol -(0+0)[/tex]

[tex]\Delta H_{rxn1} =-393.5 \ kJ/mol[/tex]

From (2), the enthalpy change is:

[tex]\Delta_{rxn2} = \Delta H^0_{fCO_2(g)} - ( \Delta H^0_{fCO(g)} + \dfrac{1}{2} \Delta H^0_{fO_2(g)})[/tex]

[tex]\Delta_{rxn2} = -393.5 \ kJ/mol -(-110.5 + \dfrac{1}{2}(0))[/tex]

[tex]\Delta_{rxn2} = -283.0 \ kJ/mol[/tex]

Subtracting (2) from (1), we get:

[tex]C(graphite) + O_{2(g)} \to CO_{2(g)} \ \ \ \Delta H_{rxn} = -393.5 \ kJ/mol}[/tex]

[tex]CO_{(g)} + \dfrac{1}{2} O_2(g) \to CO_{2(g)}} \ \ \ \Delta H _{rxn2} = -283.0 \ kJ/mol[/tex]

[tex]C(graphite) + O_{2(g)} \to CO (g) + \dfrac{1}{2}O_{2(g)} \ \ \ \Delta H_{rxn} = -110.5 \ kJ/mol[/tex]

[tex]C(graphite) + \dfrac{1}{2} O_{2(g)} \to CO (g) \ \ \ \Delta H_{rxn} = -110.5 \ kJ/mol[/tex]

The enthalpy change ΔH of the reaction = -110.5 kJ/mol

Answer:

Gallium

Explanation:

Answer:

Gallium

Explanation:

Answer:

B. Cell Type

Explanation:

that is the answer i believe

Answer:

option b- cell type

Explanation:

almost absolutely positive

Answer:

it is all of the above

Explanation:

Answer:

17.031 g/mol

Explanation:

Answer:

a. 1-fluoro-3,3,4-trimethyl-pentane.

b. 1-iodo-3-ethyl-2-methyl-hexane.

c. 1,3-dichloro-5-dimethyl-hexane.

d. 1-bromo-3-methyl-cyclopentane.

Explanation:

Hello!

In this case, according to the IUPAC rules for the listed alkyl halides, we first need to name the halogens (considering periodic order) then alkyl radicals and finally the parent chain; thus, the names are given below:

a. 1-fluoro-3,3,4-trimethyl-pentane.

b. 1-iodo-3-ethyl-2-methyl-hexane.

c. 1,3-dichloro-5-dimethyl-hexane.

d. 1-bromo-3-methyl-cyclopentane.

e. 5-chloro-1-bromo-1,1,5-trimethyl-pentane (radicals are not clear).

Best regards!

Answer:

348.9 mL of the 60% solution and 251.1 mL of the 15% solution.

Explanation:

First, we calculate how many mililiters of pure carbonic acid are there in 650 mL of a 38% solution:

Then we can express the sum of both initial solutions as:

for the volume of carbonic acid; and

For the volume of the solutions.

We now have a system of two equations and two unknowns (x is the volume of the 60% solution and y is the volume of the 15% solution).

We express x in terms of y in equation 2):

And replace x in equation 1):

Finally we calculate x using equation 2):

Answer:

there's just a lot of drama around grumpy trumpy's StOlEN eLecTiOn which are all just fake claims. all his hardcore followers are brainwashed into believing anything he says, so people (especially people who are inside the capitol) fear that trump's followers might incite violence AGAIN even after they broke into the Capitol on the 6th.

That's why they have the National Guard and thousands of people to keep the building secure during Biden's Inauguration.

hope this helps!

Answer:

9.11×10⁷ Hz

Explanation:

From the question given above the following data were obtained:

Energy (E) = 6.04×10¯²⁹ KJ

Frequency (f) =?

Next, we shall convert 6.04×10¯²⁹ KJ to joule (J). This can be obtained as follow:

1 KJ = 1000 J

Therefore,

6.04×10¯²⁹ KJ = 6.04×10¯²⁹ KJ × 1000 J / 1 KJ

6.04×10¯²⁹ KJ = 6.04×10¯²⁶ J

Thus, 6.04×10¯²⁹ KJ is equivalent to 6.04×10¯²⁶ J.

Finally, we shall determine the frequency at which the FM radio station is broadcasting as follow:

Energy (E) = 6.04×10¯²⁶ J

Planck's constant (h) = 6.63×10¯³⁴ Js

Frequency (f) =?

E = hf

6.04×10¯²⁶ = 6.63×10¯³⁴ × f

Divide both side by 6.63×10¯³⁴

f = 6.04×10¯²⁶ / 6.63×10¯³⁴

f = 9.11×10⁷ Hz

Therefore, the frequency at which the FM radio station is broadcasting is 9.11×10⁷ Hz

Answer:

[tex]MM_{mixture}=9.68g/mol[/tex]

Explanation:

Hello!

In this case, since the molar mass of gaseous mixtures can be determined via a weighted average containing the molar mass of each gas and the mole fraction (mole percent divided by 100) as shown below:

[tex]MM_{mixture}=x_{CO_2}M_{CO_2}+x_{N_2}M_{N_2}+x_{He}M_{He}[/tex]

Now, we plug in to obtain:

[tex]MM_{mixture}=0.11*44g/mol+0.053*28+0.84*4\\\\MM_{mixture}=9.68g/mol[/tex]

Best regards!

Answer:

V = 11.2L are produced

Explanation:

... Sample of 27g of pure aluminium, 3added to 333 mL of 3.0 M HCl..

Based on the chemical reaction:

2Al(s) + 6HCl(aq) → 2AlC₃(aq) + 3H₂(g)

Where 3 moles of hydrogen are produced when 6 moles of hydrochloric acid reacts with 2 moles of Al.

To solve this question, we need to determine limiting reactant converting each reactant to moles. With limiting reactant and the chemical reaction we can find moles of hydrogen and its volume at STP (T=273.15K; P=1atm), thus:

Moles Al-Molar mass: 26.98g/mol-:

27g * (1mol / 26.98g) = 1mol of Al

Moles HCl:

333mL = 0.333L * (3mol/L) = 1mol HCl

For a complete reaction of 1 mole of HCl are required:

1mol HCl * (2mol Al / 6mol HCl) = 0.333 moles of Al. As there is 1 mole of Al, Al is in excess and HCl is limiting reactant.

Moles of Hydrogen produced are:

1mol HCl * (3 moles H₂ / 6 mol HCl) = 0.5moles H₂ are produced.

Using ideal gas law:

PV = nRT

V = nRT/P

Where V is volume

n are moles: 0.5mol

R is gas constant: 0.082atmL/molK

T is absolute temperature: 273.15K

P is pressure: 1atm.

Solving for V:

V = 0.5mol*0.082atmL/molK*273.15K / 1atm

Answer:

B

Explanation:

Answer:

- P in PBr3 is +3.

- N in N2 is 0.

- As in H3AsO4 is +5.

Explanation:

Hello!

In this case, since the determination of the oxidation states is performed by using the well-known charge balances, we can proceed as shown below:

- P in PBr3: Here, bromide ions have an oxidation state of -1, so we follow:

[tex]P^xBr_3^-\\\\x-3=0\\\\x=+3[/tex]

Thus, the oxidation state is +3.

- N in N2: Here, since nitrogen is bonding with nitrogen and it is neutral, we infer its oxidation state is 0.

- As in H3AsO4: Here, oxygen is -2 and hydrogen +1, so we follow:

[tex]H_3^+As^xO_4^{-2}\\\\3+x-8=0\\\\x=8-3\\\\x=5[/tex]

Thus, the oxidation state is +5.

Best regards!

Answer:  Approximately [tex]0.267 \frac{\text{J}}{\text{g}^{\circ}\text{C}}[/tex]

===================================================

Work Shown:

We have the following variables

note: the symbol [tex]\Delta[/tex] is the uppercase Greek letter delta. It represents the difference or change in a value.

Apply those values into the formula below. Solve for c.

[tex]Q = m*c*\Delta t\\\\714 = 52*c*51.5\\\\714 = 52*51.5*c\\\\714 = 2678*c\\\\2678*c = 714\\\\c = \frac{714}{2678}\\\\c \approx 0.26661687826737\\\\c \approx 0.267\\\\[/tex]

The specific heat of the unknown metal is roughly [tex]0.267 \frac{\text{J}}{\text{g}^{\circ}\text{C}}[/tex]

Answer: $6,600[tex]\alpha \sqrt[n]{x} \lim_{n \to \infty} a_n[/tex]

Explanation:

Answer:

The First Battle of Panipat was fought between the invading forces of Babur and the Lodi Empire, which took place on 21 April 1526 in North India. It marked the beginning of the Mughal Empire. This was one of the earliest battles involving gunpowderfirearms and field artillery.

For you to write an essay on  Panipat first bring into view

Then write a brief story using this format.

In writing this essay, you must consider the structure and content of the essay or short note

For structure kick off with

In conclusion, for the content of the short note you can consider this,

Panipat fought its first battle on 21 April 1526 in the North of India, This battle was fought between the invading powers of Babur and the Lodi Empire.

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Answer: 1.695x10^24 atoms

Explanation: 0.938 moles CO2 = 0.938xavogadro number molecules CO2, so number of atoms = 3x0.938x6.02214076x10^23 = 1.695x10^24

Answer:

wouldnt they still have the same density?

Explanation:

for example, the density of iron is 7.874 g/cm^3 no matter how big the piece is. I hope this helps.

Answer:

It does not matter where the sample of water came from or how it was prepared. Its composition, like that of every other compound, is fixed.