The molecular mass Of a compound of carbon and hydrogen is 42. What is the empirical formula of the compound

Answer:

[tex]Q=50,849.25J\\\\Q=12,153.3cal[/tex]

Explanation:

Hello!

In this case, given the mass, temperature change and specific heat, it is possible to compute the required heat in joules as shown below:

[tex]Q=mC(T_2-T_1)\\\\Q=75.0g*0.449\frac{J}{g\°C}(1535\°C-25\°C)\\\\Q=50,849.25J[/tex]

Now, since 1 cal =4.184 J, this result in calories is:

[tex]Q=50,849.25J*\frac{1cal}{4.184J}\\\\Q=12,153.3cal[/tex]

Best regards!

Answer:

0.322 moles

Explanation:

Given Mass, m = 100 g

The molar mass of Ca₃(PO₄)₂

Ca₃(PO₄)₂ = (3 x 40.08) + (2 x 30.97) + (8 x 16.00)

M = 310.18 g/mol

Let there are n number of moles,

n = given mass/ molar mass

[tex]n=\dfrac{100 }{310.18 }\\\\n=0.322\ mol[/tex]

So, there are 0.322 moles.

Answer:

c. 2s2 2p5

Explanation:

2s2 2p5 has 7 valence electrons and only needs one electron to complete the octet.  This element will be the most electronegative.

Answer:

yes it does

Explanation:

I searched it up

Answer:

Third Option

Explanation:

Im Elon Musk

Answer:

5 grams remain

Explanation:

A = A₀exp(-k·t) 80g·exp(-0.00238yr⁻¹)(1165yrs) = 80g x 0.063 = 5 grams remain

Answer:

S (Sulfur)

Explanation:

Electronegativity increases as you go up and right on the periodic table, excluding the noble gases on the rightmost column because they have zero electronegativity. S is the farthest right out of all the elements listed in the question excluding Ne.

Answer:

Iron (II) sulfide and oxygen

Answer:

6.60 moles of ZnO have 537,108 g.

Explanation:

Answer:

N2 (nitrogen) O3 (ozone) CaO (calcium oxide) C6H12O6 (glucose, a type of sugar)

Answer:

dihydrogen monoxide

dinitrogen monosulfide

Answer:

Big Chill

Explanation:

Answer:

it's big chill or big freeze

Answer:

3.94 L

Explanation:

From the question given above, the following data were obtained:

Mass of O₂ = 5.62 g

Volume of O₂ =?

Next, we shall determine the number of mole present in 5.62 g of O₂. This can be obtained as follow:

Mass of O₂ = 5.62 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mole of O₂ =?

Mole = mass / molar mass

Mole of O₂ = 5.62 / 32

Mole of O₂ = 0.176 mole

Finally, we shall determine the volume of 5.62 g (i.e 0.176 mole) of O₂ at STP. This can be obtained as follow:

1 mole of O₂ occupied 22.4 L at STP.

Therefore, 0.176 mole of O₂ will occupy = 0.176 × 22.4 = 3.94 L at STP.

Thus 5.62 g (i.e 0.176 mole) of O₂ occupied 3.94 L at STP

Answer:

0.080 M

Explanation:

From the question given above, the following data were obtained:

Concentration of H₂SO₃ = 0.080 M

Concentration of SO₃²¯ =…?

The concentration of SO₃²¯, can be obtained as follow:

H₂SO₃ (aq) <==> 2H⁺ (aq) + SO₃²¯ (aq)

From the balanced equation for above,

1 mole of H₂SO₃ produced 1 mole of SO₃²¯.

Therefore, 0.080 M H₂SO₃ will also produce 0.080 M SO₃²¯.

Thus, the concentration of SO₃²¯ is 0.080 M

Answer:

hardness im sure

Explanation:

Answer:

B.

A.

D.

A.

C.

Hope it helps..

Have a great day : )

Answer:

★ Molecular geometry is described by VSEPR theory, which basically states that electron pairs around a central atom will repel each other, and get as far apart as possible, in three dimensions.

Explanation:

Hope you have a great day :)

Answer:

Carbon and hydrogen

Explanation:

C = Carbon

H = Hydrogen

Answer:

can u mark me brainliest

Explanation:

Following the Law

Matter cannot be created or destroyed in chemical reactions. This is the law of conservation of mass. In every chemical reaction, the same mass of matter must end up in the products as started in the reactants. Balanced chemical equations show that mass is conserved in chemical reactions

Answer:

Wheat!

Explanation:

Corn ranks first, Wheat, then Rice!

1100 mmHg

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Gas Laws

Stoichiometry

Step 1: Define

[Given] 1.5 atm

[Solve] mmHg

Step 2: Identify Conversions

1 atm = 760 mmHg

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

1140 mmHg ≈ 1100 mmHg

P₂ = 0.67 atm

Math

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Chemistry

Gas Laws

Boyle's Law: P₁V₁ = P₂V₂

Step 1: Define

[Given] P₁ = 2.02 atm

[Given] V₁ = 4.0 L

[Given] V₂ = 12.0 L

[Solve] P₂

Step 2: Solve

Step 3: Check

Follow sig fig rules and round. We are given 2 sig figs as our smallest.

0.673333 atm ≈ 0.67 atm

Answer: 34 gram of chromium produced from 50g of [tex]Cr_2O_3[/tex]

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} Cr_2O_3=\frac{50g}{152g/mol}=0.33moles[/tex]

The decomposition of [tex]Cr_2O_3[/tex] follows the equation :

[tex]2Cr_2O_3\rightarrow 4Cr+3O_2[/tex]  

According to stoichiometry :

2 moles of [tex]Cr_2O_3[/tex] produce = 4 moles of [tex]Cr[/tex]

Thus 0.33 moles of [tex]Cr_2O_3[/tex] will produce=[tex]\frac{4}{2}\times 0.33=0.66moles[/tex]  of [tex]Cr[/tex]

Mass of [tex]Cr=moles\times {\text {Molar mass}}=0.66moles\times 52g/mol=34g[/tex]

Thus 34 gram of chromium produced from 50g of [tex]Cr_2O_3[/tex]

Explanation:

The equation of the reaction is given as;

Be + 2HCl → BeCl2 + H2

What is the mass of beryllium required to produce 25.0g of beryllium chloride?

1 mol of Be produces 1 mol of BeCl2

Converting to mass;

Mass = Molar mass  *  Number of moles

9.01g of Be produces 79.92g of BeCl2

xg of Be produces 25g of BeCl2

Solving for x;

x = 25 * 9.01 / 79.92

x = 2.82 g

What is the mass of hydrochloric acid required to produce 25.0g of beryllium chloride? g

Converting 25.0g of beryllium chloride to moles;

Number of moles = Mass / Molar mass

Number of moles = 25 / 79.92 = 0.3128 mol

2 mol of HCl produces 1 mol of BeCl2

x mol of HCl would produce 0.3128 mol of BeCl2

solving for x;

x = 0.3128 * 2 = 0.6256 mol

Converting to mass;

Mass = 0.6256 * 36.5 = 22.83 g

What is the mass of hydrogen gas produced when 25.0g of beryllium chloride is also produced? g

25g of BeCl2 = 0.3128 mol of BeCl2

From the equation;

1 mol of H2 is produced alongside 1 mol of BeCl2

This means;

0.3128 mol of H2 would also be produced alongside 0.3128 mol of BeCl2

Mass = Number of moles * Molar mass

Mass = 0.3128mol * 2.0159 g/mol = 0.6306 g

Answer:

kay, so u didn't gave the list but I guess I'm right...bcz of some words you wrote below..

yeah,

so your answer is GERMINATION

Answer:

Cementation

Erosion

Explanation:

Cementation refers to hardening and welding of those sediments which are formed from preexisting rock fragments as a result of the precipitation of mineral matter in the spaces between the pore. The last stage in the formation of a sedimentary rock is Cementation.

Erosion is the geological process in which soil, rock, or dissolved material are worn away and transported from one location to another location by natural forces such as wind or water.

The two parts of sedimentary rock formation which include the breakdown and carrying away of existing rock are cementation and erosion.

Answer: The answer is A: Cementation and Erosion

Explanation: Hope this helps! happy new year

Explanation:

Here is a history of questions and answers processed by "Ask the Physicist!". If you like my answer, please consider making a donation to help support this service.

If there is a link to a previously answered question, be patient. Since the files containing the older answers are rather large, it takes some time (maybe as much as 15 seconds or so) to find the appropriate bookmark.

Answer: 445.15 kJ of heat could be obtained by the combustion of 8.0 grams [tex]CH_4[/tex]

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} CH_4=\frac{8.0g}{16.04g/mol}=0.50moles[/tex]

The given thermochemical equation is:

[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)[/tex]  [tex]\Delta H=890.3kJ[/tex]

According to stoichiometry :

1 mole of [tex]CH_4[/tex] on combustion produce heat = 890.3 kJ

Thus 0.50 moles of [tex]CH_4[/tex] on combustion produce heat =[tex]\frac{890.3}{1}\times 0.50=445.15kJ[/tex]

Thus 445.15 kJ of heat could be obtained by the combustion of 8.0 grams [tex]CH_4[/tex]