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andreyandreev [35.5K]
3 years ago
14

Which must be the same when comparing 1 mol of oxygen gas, O2, with 1 mol of carbon monoxide gas, CO?

Chemistry
1 answer:
Julli [10]3 years ago
7 0

Answer:

The number of molecules.

Explanation:

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Hi, someone knows how to balance the following equation, and if you can thank you if you can explain how that is done:
kari74 [83]
To begin with, the equation given is not correct.
Correct equation is : CaCO3 + HCl ---> CaCl2 + H2O + CO2
It's CaCl2 not CaCl because Ca has a valency of 2 

    LHS                          RHS
CaCO3 + HCl ---> CaCl2 + H2O + CO2
First of all, to balance the equation you must look at the number of atoms on each side of the equation. 
we have 2 H on the RHS and 1 H on the LHS. So, we put a 2 on the LHS

CaCO3 + 2HCl ---> CaCl2 + H2O + CO2
Check for the LHS: 1 Ca, 1 C, 3 O, 2 H & 2 Cl on the LHS
Now check for the RHS: 1 Ca, 2 Cl, 2 H, 1 C & 3 O

Hope it helped!
7 0
3 years ago
What is the concentration (M) of CH3OH in a solution prepared by dissolving 34.4 g of CH3OH in sufficient water to give exactly
bonufazy [111]

Answer:

4.67M

Explanation:

The concentration of methanol (CH3OH) can be calculated using the following:

Molarity (M) = number of moles(n)/volume(v)

However, mole is not given. It can be obtained by using:

Mole = mass / molar mass

Where; mass = 34.4g

Molar mass (MM) of CH3OH is:

= 12 + 1(3) + 16 + 1

= 12 + 3 + 17

= 32g/mol

mole = 34.4/32

mole = 1.075mol

The volume needs to be converted to L by dividing by 1000

230mL = 230/1000

= 0.230L

Molarity = mol/volume

Molarity = 1.075/0.230

Molarity = 4.6739

Molarity = 4.67M

The concentration of CH3OH in solution is 4.67M

5 0
3 years ago
What mass of water was produced if 350.0 L of carbon dioxide were made at STP? C3H8 (g) + 5O2(g) → 2CO2 (g) + 4H2O (g)
madam [21]

700 L of water was produced if 350.0 L of carbon dioxide were made at STP.

The quantitative relationship (ratio) between reactants and products in a chemical reaction that produces gases is known as gas stoichiometry. When the created gases are presumed to be ideal and their temperature, pressure, and volume are all known, gas stoichiometry is applicable.

The ideal gas equation is PV=nRT, where n is the number of moles and R is the gas constant, P is the pressure measured in atmospheres (atm), V is the volume measured in liters (L), and

Calculations based on stoichiometry assist scientists and engineers who work in the business world in estimating the number of items they will make using a particular process. They can also assist in determining if a product will be economical to produce.

Reduced growth, reproduction, and survivability for the consumer are typically the results of a significant stoichiometric imbalance between the primary producer and consumer.

To know more about stoichiometry refer to:  brainly.com/question/9743981

#SPJ1

4 0
2 years ago
For the following reaction, KcKc = 255 at 1000 KK.
bonufazy [111]

Answer :

The equilibrium concentration of CO is, 0.016 M

The equilibrium concentration of Cl₂ is, 0.034 M

The equilibrium concentration of COCl₂ is, 0.139 M

Explanation :

The given chemical reaction is:

                           CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

Initial conc.      0.1550      0.173           0

At eqm.          (0.1550-x)  (0.173-x)         x

As we are given:

K_c=255

The expression for equilibrium constant is:

K_c=\frac{[COCl_2]}{[CO][Cl_2]}

Now put all the given values in this expression, we get:

255=\frac{(x)}{(0.1550-x)\times (0.173-x)}

x = 0.139 and x = 0.193

We are neglecting value of x = 0.193 because equilibrium concentration can not be more than initial concentration.

Thus, we are taking value of x = 0.139

The equilibrium concentration of CO = (0.1550-x) = (0.1550-0.139) = 0.016 M

The equilibrium concentration of Cl₂ = (0.173-x) = (0.173-0.139) = 0.034 M

The equilibrium concentration of COCl₂ = x = 0.139 M

5 0
3 years ago
Which of the following is not true of a scientific theory?
Furkat [3]

Answer: a,b

Explanation: because a scientific theory is something that been thought of and tested multiple times.

3 0
3 years ago
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