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3 years ago

Chlorine (Cl) forms a salt when it is combined with a metal. This element belongs in _____.

Chemistry

1 answer:

ella [17]3 years ago

8 0

Chlorine (Cl) forms a salt when it is combined with a metal. This element belongs in halogens. GROUP 17

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The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of B

loris [4]

I believe here is the right question, so will just ignore the rest of the junk information from the previous message

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.944 g of H2O(g). Insert subscripts below to appropriately display the empirical formula of BHT

Answer:

$C_{15}H_{24}0$

Explanation:

A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.994 g of H2O(g).

If all the carbon in BHT is present in $CO_2$ and also, all the hydrogen in BHT is present in $H_2O$ , Then we can determine for the corresponding numbers of moles of Carbon(C) and Hydrogen (H) respectively as:

moles of $CO_2$ = $8.990 g*(\frac{1mole}{44.01g})$

= 0.2043 moles

∴ moles of C = 0.2043 moles

moles of $H_2O$ = $2.944 g *(\frac{1mole}{18.01g} )$

= 0.1635 moles

∴ moles of H = 2 × 0.1635 moles

= 0.327 moles

Since number of moles= $\frac{mass}{molarmass}$

number of moles of H = 0.327 moles

molar mass of H = 1.008 g/mol

∴ mass of H in the sample = 0.327 moles × 1.008 g/mol

= 0.329616g

mass of C in the sample can be calculated as = 0.2043 moles × $(\frac{12.01g}{1 mole} )$

= 2.453643 g

mass of C+H in the sample = 2.453643g + 0.329616g

= 2.783259 g

mass of O can be calculated as = 3.001 g - 2.783259 g

= 0.217741 g

∴ moles of O = 0.217741g × $(\frac{1mole}{16.0g})$

= 0.0136 moles

Now, since; we've gotten our data, we can now proceed to calculate for the empirical formula.

C H O

0.2043 0.327 0.0136

Dividing by the least number (0.0136) , we have :

$\frac{0.2043}{0.0136}$ $\frac{0.327}{0.0136}$ $\frac{0.0136}{0.0136}$

15.02 24.04 1

≅

15 24 1

Therefore, the empirical formula would be : $C_{15}H_{24}0$

7 0

2 years ago

For the reaction COCl2(g)⇌CO(g)+Cl2(g), K= 2.19×10−10 at 373 K

vladimir1956 [14]

Answer:

D) The equilibrium lies far to the left

Explanation:

According to the law of mass action, the equilibrium constant K for the reaction at 373K can be calculated as follows:

K = $\frac{[CO][Cl_{2}]}{[COCl_{2}]}$ = 2.19×10^{-10}

([X] means = concentration of X)

This means that in the equilibrium the concentration of the reactant (that is in the denominator) will be much higher (around 10^{10} fold) than the concentrations of the products (that are in the numerator), and this means that the equilibrium lies far to the left (to the reactants side) as very small amount of product is being formed.

6 0

3 years ago

What molecule is removed during dehydration synthesis?.

nignag [31]

Answer: two hydrogen atoms and one oxygen atom are removed during dehydration synthesis.

Hope this helps!

4 0

2 years ago

5) Calculate the molality of 0.210 mol of KBr dissolved in 0.075kg pure water?

Margaret [11]

Answer:

$\boxed {\boxed {\sf 2.8 \ m }}$