All categories

9 months ago

Can you discount alcohol in Texas?.

1 answer:

7 0

According to TABC laws on alcohol use, it is illegal to sell alcohol in many places in texas thus no you can't discount alcohol in Texas.

Alcohol is prohibited in many parts of Texas thus anyone under the age of 21 may not be sold alcohol, including adults. Beer and wine may be purchased from licensed establishments Monday through Friday from 7:00 a.m. to midnight. On Saturday, they may sell them from 7:00 a.m. until 1:00 p.m. Additionally, they may sell on Sundays from noon to midnight. Only liquor stores, nevertheless, are allowed to sell distilled alcohol. On Sundays before noon, restaurants are allowed to serve alcoholic beverages with food. On the other hand, it is prohibited to sell any alcoholic beverages with a volume of more than 17% on Sunday (ABV). On Thanksgiving Day, New Year's Day, or Christmas Day, no establishment may offer any alcoholic beverages. Alcohol from a single producer cannot be sold by a restaurant or other retailer. And alcohol cannot be sold by food trucks.

To learn more about alcohol use in texas please click on the given link: brainly.com/question/28286273

#SPJ4

You might be interested in

Which of these statement(s) is/are true about a balloon filled with 1.00 mol N2 (g) at STP?

lara [203]

Answer:

I. The balloon has a volume of 22.4L

III. The balloon contains 6.022x10^23 molecules.

Explanation:

At stp, it has been proven that 1mole of a gas occupy 22.4L.

Therefore, option (i) is correct.

The molar mass N2 = 14.01 x 2 = 28.02g/mol

Number of mole of N2 = 1 mole

Mass of N2 =..?

Mass = mole x molar Mass

Mass of N2 = 1 x 28.02 = 28.02g.

The mass content of the balloon is 28.02g, therefore, option (ii) is wrong.

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10^23 molecules. This implies that 1 mole of N2 also contains 6.02x10^23 molecules

Therefore, option (iii) is correct.

The correct options to the question are:

Option i and option iii

7 0

2 years ago

Read 2 more answers

I need help with this assignment

yuradex [85]

The answers for the following sums is given below.

1. $Pd_{2}H_{2}$

2. $C_{2} H_{6}$

3. $C_{2} H_{2} O_{2} Cl_{2}$

4. $C_{3}Cl_{3} N_{3}$

5. $Tl_{2 } C_{4} H_{4}O_{6}$

6. $C_{8}H_{8}$

7. $N_{2}O_{5}$

8. $P_{4}O_{6}$

9. $C_{4}H_{8} O_{2}$

Explanation:

1.Given:

Molar mass=216.8g

Molecular formula=Pd $H_{2}$

we know;

Molecular formula=n(Empirical formula)

molecular weight of palladium(Pd)=106.4u

molecular weight of hydrogen(H)=1u

Molar mass of Pd $H_{2}$ :

Pd=106.4×1=106.4u

H=1×2=2

molar mass of Pd $H_{2}$ =106.4+2=108.4

n= $\frac{216.8}{106.4}$

n=2

Molecular formula=2(Pd $H_{2}$ )

Molecular formula= $Pd_{2}H_{2}$

Therefore the molecular formula of the compound is $Pd_{2}H_{2}$

2. Given:

Molar mass=30.0g

Molecular formula= $CH_{3}$

we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

Molar mass of $CH_{3}$ :

C=12.01 × 1 = 12.01u

H=1 × 3 = 3u

molar mass of $CH_{3}$ =12.01 + 3 =15.01u

n= $\frac{30.0}{15.01}$

n=2

Molecular formula=2( $CH_{3}$ )

Molecular formula= $C_{2} H_{6}$

Therefore the molecular formula of the compound is $C_{2} H_{6}$

3. Given:

Molar mass=129g

Molecular formula=CHOCl

we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

molecular weight of oxygen(O)=16.00u

molecular weight of chlorie(Cl)=35.5u

Molar mass of CHOCl:

C=12.01 × 1 = 12.01u

H=1 × 1 = 1u

O=16.00×1=16.00u

Cl=35.5×1=35.5u

molar mass of CHOCl=12.01+1+16.00+35.5=64.5u

n= $\frac{129}{64.5}$

n=2

Molecular formula=2(CHOCl)

Molecular formula= $C_{2} H_{2} O_{2} Cl_{2}$

Therefore the molecular formula of the compound is $C_{2} H_{2} O_{2} Cl_{2}$

5. Given:

Molar mass=577g

Molecular formula= $TlC_{2} H_{2}O_{3}$

we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of Thallium(Tl)=204.3u

molecular weight of hydrogen(H)=1u

molecular weight of oxygen(O)=16.00u

Molar mass of $TlC_{2} H_{2}O_{3}$ :

C=12.01 × 2= 24.02u

Tl=204.3×1=204.3u

H=1×2=2u

O=16.00×3=48.00

molar mass of $TlC_{2} H_{2}O_{3}$ =204.3+24.02+1+48.00=278.32u

n= $\frac{577}{278.32}$

n=2

Molecular formula=2 ( $TlC_{2} H_{2}O_{3}$ )

Molecular formula= $Tl_{2 } C_{4} H_{4}O_{6}$

Therefore the molecular formula of the compound is $Tl_{2 } C_{4} H_{4}O_{6}$

4. Molar mass=184.5g

Molecular formula=CClN

we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of Nitrogen(N)=14u

molecular weight of chlorine(Cl)=35.5u

Molar mass of CClN:

C=12.01 × 1 = 12.01u

N=1×14=14U

Cl=35.5×1=35.5u

molar mass of CClN=12.01+14+35.5=61.5u

n= $\frac{184.5}{61.5}$

n=3

Molecular formula=3 (CClN)

Molecular formula= $C_{3}Cl_{3} N_{3}$

Therefore the molecular formula of the compound is $C_{3}Cl_{3} N_{3}$

6. For the table refer the attached file.

Simplest ratio of elements:

Carbon=8

Hydrogen=8

Empirical formula= $C_{8}H_{8}$

Molecular formula = $C_{8}H_{8}$

Molar mass of $C_{8}H_{8}$ :

molecular weight of carbon=12.04u

molecular weight of hydrogen=1u

C=8×12.01=96.08u

H=1×8=8u

molar mass of $C_{8}H_{8}$ =96.08+8=104.08u

n=104.08÷78.0

n=1

Molecular formula = n(Empirical formula)

Molecular formula = 1( $C_{8}H_{8}$ )

Molecular formula = $C_{8}H_{8}$

Therefore the molecular formula of a compound is $C_{8}H_{8}$

7. Given:

mass of oxide of nitrogen=108g

mass of nitrogen=4.02g

mass of oxygen=11.48g

moles of nitrogen= $\frac{4.04}{14.01}$ = 0.289 moles

moles of oxygen= $\frac{11.46}{15.999}$ =0.716 moles

We divide through by the lowest molar quantity to give an empirical formula of $N_{2} O{5}$ .

Now the molecular formula is multiple of the empirical formula.

So,

108 = n × (2×14.01 + 5×15.999)

Clearly,n=1, and the molecular formula is $N_{2}O_{5}$ .

8.For the table refer the attached file.

Simplest ratio of elements:

Phosphorus=2

Oxygen=3

We know;

Empirical formula= $P_{2} O_{3}$

molecular formula= 2(Empirical formula)

Molecular formula =2( $P_{2} O_{3}$ )

Molecular formula = $P_{4}O_{6}$

Therefore the molecular formula of the compound is $P_{4}O_{6}$

9. For the table refer the attached file.

Simplest ratio of elements:

Carbon=2

Hydrogen=9

Oxygen=2

We know;

Empirical formula = $C_{2} H_{4} O$

Molecular formula = 2(Empirical formula)

Molecular formula =2( $C_{2} H_{4} O$ )

Molecular formula = $C_{4}H_{8} O_{2}$

Therefore the molecular formula of the compound is $C_{4}H_{8} O_{2}$

4 0

2 years ago

Which pH value represents a solution with the lowest OHM ion concentration?

Arlecino [84]

pH value 1 represents a solution with the lowest OH⁻ion concentration.

Explanation:

pH is given by the expression as the negative logarithm to the base 10 of the concentration of hydrogen ions.

pH = -log₁₀[H⁺]

If the pH is lower than 7, pH < 7 then it is acidic

If the pH = 7, then it is neutral

If the pH > 7, then it is basic

If pH is 1 then the solution is showing mostly acidic character,which is least basic in its character.

So if the pH is 1, which is most acidic and least basic solution that is lowest OH⁻ ion concentration.

7 0

2 years ago

If 0.250 mol of hydrogen gas occupies 0.333L at 20.0c what is the pressure in atmospheres?

maria [59]

Answer: The pressure will be 18.05 atm.

Explanation: Expression for ideal gas equation is :

$PV=nRT$

where,

P = Pressure of the gas = ? atm

V = Volume of the gas = 0.333L

n = Number of moles of gas = 0.250 moles

R = Universal gas constant = $0.082057\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the gas = 20°C = (273 + 20)K = 293K

Putting values in above equation, we get:

$P\times 0.333L=0.250mol\times 0.082057\text{ L atm }mol^{-1}K^{-1}\times 293K$

P = 18.05 atm

8 0

3 years ago

An 80.0g sample of an unknown metal is at an initial temperature of 55.5oC. Afer 540 J of energy is absorbed by the metal, the t

Lunna [17]

Answer:

Specific heat of metal = 0.26 j/g.°C

Explanation:

Given data:

Mass of sample = 80.0 g

Initial temperature = 55.5 °C

Final temperature = 81.75 °C

Amount of heat absorbed = 540 j

Specific heat of metal = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 81.75 °C - 55.5 °C

ΔT = 26.25 °C

540 j = 80 g × c × 26.25 °C

540 j = 2100 g.°C× c

540 j / 2100 g.°C = c

c = 0.26 j/g.°C

7 0

2 years ago