A) fluorite B) orthoclase C) apatite D) gypsum

What is a hot spot? ( Must be in your own words) Pleaseeee Hurryy

Answer:

A hot spot is an area on Earth over a mantle plume or an area under the rocky outer layer of Earth, called the crust, where magma is hotter than surrounding magma. The magma plume causes melting and thinning of the rocky crust and widespread volcanic activity.

Hope this is what you mean be hot spot!

I hope this helps you!

Have a great day

7 0

3 years ago

What is the wavelength of light with 2.89 x 10-19 J of energy? (The speed of

Alika [10]

Let's see

$\\ \rm\rightarrowtail E=hv$

$\\ \rm\rightarrowtail E=\dfrac{hc}{\lambda}$

lambda is wavelength
h is Planck constant
c is velocity of sound in air
E in energy

$\\ \rm\rightarrowtail \lambda=\dfrac{hc}{E}$

$\\ \rm\rightarrowtail \lambda=\dfrac{6.626\times 10^{-34}\times 3\times 10^8}{2.89\times 10^{-19}}$

$\\ \rm\rightarrowtail \lambda=6.878\times 10^{-7}m$

$\\ \rm\rightarrowtail \lambda=68.8\mu m$

3 0

2 years ago

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

4 years ago

Calculate the enthalpy of formation of butane, C4H10, using the balanced chemical

melisa1 [442]

Answer:

-125.4

Explanation:

Target equation is 4C(s) + 5H2(g) = C4H10

These are the data equations for enthalpy of combustion

C(s) + O2(g) =O2(g) -393.5 kJ/mol * 4
H2(g) + ½O2(g) =H20(l) = 285.8 kJ/mol * 5
2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse

To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3, so...

Calculate 4(-393.5) + 5(-285.8) + 2877.6 and you should get the answer.

7 0

3 years ago

3. In an experiment it was found that 40.0cm of 0.2M sodium hydroxide solution just neutralized 0.2g

hjlf

The relative molecular mass of acid A : 50 g/mol

<h3>Further explanation</h3>

Given

40.0 cm³(40 ml) of 0.2M sodium hydroxide

0.2g of a dibasic acid

Required

the relative molecular mass of acid A

Solution

Titration formula

M₁V₁n₁=M₂V₂n₂

n=acid/base valence(number of H⁺/OH⁻)

NaOH ⇒ n = 1

Dibasic acid = diprotic acid (such as H₂SO₄)⇒ n = 2

mol = M x V

Input the value in the formula :(1 = NaOH, 2=dibasic acid)

0.2 x 40 x 1 = M₂ x V₂ x 2

M₂ x V₂ = 4 mlmol = 4.10⁻³ mol ⇒ mol of Acid A

The relative molecular mass of acid A (M) :

$\tt M_A=\dfrac{mass }{mol}=\dfrac{0.2~g}{4.10^{-3}}=50~g/mol$

5 0

3 years ago