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4 years ago

A chemical reaction produced 177.3 grams of chlorine has (Cl²). What was the volume of the gas?

1 answer:

3 0

Since there is so little information given, I will assume that we are at STP and i can use the conversion factor at STP--->> 22.4 Liters= 1 mol of gas

before we use this conversion, we need to convert the grams to moles using the molar mass of the molecule.

molar mass of Cl₂= 35.5 x 2= 71.0 g/ mol

177.3 g (1 mol/ 71.0 g)= 2.50 mol Cl₂

then we use the conversion to get the volume

2.50 mol Cl₂ (22.4 Liters/ 1 mol)= 55.9 Liters

You might be interested in

Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following

bearhunter [10]

Answer:

$210.7~g~MgCO_3$

Explanation:

We have to start with the reaction:

$MgCO_3~->~MgO~+~CO_2$

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the number of moles that we have in the 110.0 g of carbon dioxide, to this, we have to know the atomic mass of each atom:

C: 12 g/mol

O: 16 g/mol

Mg: 23.3 g/mol

If we take into account the number of atoms in the formula, we can calculate the molar mass of carbon dioxide:

$(12*1)+(16*2)=44~g/mol$

In other words: $1~mol~CO_2=~44~g~CO_2$ . With this in mind, we can calculate the moles:

$110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2$

Now, the molar ratio between carbon dioxide and magnesium carbonate is 1:1, so:

$2.5~mol~CO_2=2.5~mol~MgCO_3$

With the molar mass of $MgCO_3$ ( $(23.3*1)+(12*1)+(16*3)=84.3~g/mol$ . With this in mind, we can calculate the grams of magnesium carbonate:

$2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3$

I hope it helps!

8 0

3 years ago

I want to know the steps.

Artyom0805 [142]

The answer for the following problem is described below.

Therefore the standard enthalpy of combustion is -2800 kJ

Explanation:

Given:

enthalpy of combustion of glucose(Δ $H_{f}$ of $C_{6}H_{12} O_{6}$ ) =-1275.0

enthalpy of combustion of oxygen(Δ $H_{f}$ of $O_{2}$ ) = zero

enthalpy of combustion of carbon dioxide(Δ $H_{f}$ of $CO_{2}$ ) = -393.5

enthalpy of combustion of water(Δ $H_{f}$ of $H_{2} O$ ) = -285.8

To solve :

standard enthalpy of combustion

We know;

Δ $H_{f}$ = ∈Δ $H_{f}$ (products) - ∈Δ $H_{f}$ (reactants)

$C_{6}H_{12} O_{6}$ (s) +6 $O_{2}$ (g) → 6 $CO_{2}$ (g)+ 6 $H_{2} O$ (l)

Δ $H_{f}$ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

Δ $H_{f}$ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

Δ $H_{f}$ = 6 (-393.5) + 6(-285.8) - 0 + 1275

Δ $H_{f}$ = -2361 - 1714 - 0 + 1275

Δ $H_{f}$ =-2800 kJ

Therefore the standard enthalpy of combustion is -2800 kJ

7 0

3 years ago

Can anyone check my work and see if it is correct? If not, may someone help me?

shusha [124]

It looks all correct to me, great job!

6 0

3 years ago

The temperature at which a liquid becomes a solid is the:

Olenka [21]

The answer is "c" because solidification or freezing is the term used for the process in which a liquid becomes a solid. Freezing is an exothermic process that also is an example of a phase transition

3 0

3 years ago

Baking soda and antacid tablets are examples of two common__________

attashe74 [19]

Bases

A base is a substance that dissociates into more hydroxide ions (-OH-) when dissolved in water. Bases are also good proton acceptors. Bases, therefore, reduce the number of H+ and increase OH- hence raising the pH of the solution.

B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)

Explanation:

Other properties of bases is that they are bitter to the taste and they feel slippery when touched. Strong bases are nonthlese very corrosive like acids. Bases turn red litmus paper blue. Most alkali hydroxides such as NaOH are bases.

Learn More:

For more on bases check out;

brainly.com/question/12574229

brainly.com/question/2015251

#LearnWithBrainly

7 0

3 years ago