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Natali [406]
2 years ago
7

So I lost my sense of taste yesterday.

Chemistry
2 answers:
Mama L [17]2 years ago
4 0

Answer:

I'm unaware of what but maybe hot sauce?

Explanation:

Karolina [17]2 years ago
4 0

Answer:

all

Explanation:

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Solid sodium azide (NaN3) produces solid sodium and nitrogen gas. How many grams of sodium azide are needed to yield a volume of
ch4aika [34]

Answer:

52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P*V = n*R*T

In this case, the balanced reaction is:

2 NaN₃ → 2 Na + 3 N₂

You know the following about N₂:

  • P= 1.10 atm
  • V= 26.5 L
  • n=?
  • R=0.082057 \frac{atm*L}{mol*K}
  • T= 295 K

Replacing in the equation for ideal gas:

1.10 atm* 26.5 L= n* 0.082057 \frac{atm*L}{mol*K}*295 K

Solving:

n=\frac{1.10 atm*26.5 L}{0.082057 \frac{atm*L}{mol*K} *295K}

n= 1.2 moles

Now, the following rule of three can be applied: if 3 moles of N₂ are produced by stoichiometry of the reaction from 2 moles of NaN₃, 1.2 moles of N₂ are produced from how many moles of NaN₃?

moles of NaN_{3}=\frac{1.2 molesofN_{2} *2 molesofNaN_{3} }{3 molesofN_{2} }

moles of NaN₃= 0.8

Since the molar mass of sodium azide is 65.01 g / mol, then one last rule of three applies: if 1 mol has 65.01 grams of NaN₃, 0.8 mol how much mass does it have?

mass of NaN_{3} =\frac{0.8 mol*65.01 grams}{1 mol}

mass of NaN₃=52.008 grams

<u><em>52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.</em></u>

6 0
3 years ago
Read 2 more answers
Can somebody help me I’ve been a helping hand all day and nobody helping like does the app work or I need to delete.
Liula [17]

I believe that #1 is the lie, but I'm not great at this subject.

4 0
3 years ago
A chemist combines ethane and chlorine with the goal of producing chloroethane which other product could be reasonably expected
lesya692 [45]
<span>So when the chemist combines Ethane (CH3CH3) and Chlorine (Cl2) with the intention of producing Chloroethane (CH3CH2Cl), the other product that's formed in this reaction is 1,2-dichloroethane (ClCH2CH2Cl) also called as Ethylene dichloride with molecular weight of 98.954 g/mol. This is a colorless oily flammable substance that weighs heaver when vaporized.</span>
5 0
3 years ago
Read 2 more answers
A volume of 80.0 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. I
murzikaleks [220]

Answer:

The mass of the steel bar is 26.833 grams

Explanation:

<u>Step 1: </u>data given

ΣQ gained = ΣQ lost

Q=m*C*ΔT

with m = mass in grams

with C= specific heat capacity ( in J/(g°C))

with ΔT = change in temperature = T2-T1

Qsteel = Qwater

msteel * Csteel * (T2steel - T1steel) = mwater * Cwater * (T2water - T1water)

Mass of steel = TO BE  DETERMINED

mass of water =⇒ since 1mL = 1g : 80 mL = 80g

Csteel =0.452 J/(g °C

Cwater = 4.18 J/(g °C

initial temperature steel T1 : 2 °C

final temperature steel T2 = 21.3 °C

initial temperature water T1 =22 °C

final temperature water T2 = 21.3 °C

<u>Step 2:</u> Calculate mass of steel

msteel * Csteel * (T2steel - T1steel) = mwater * Cwater * (T2water - T1water)

msteel * 0.452 *(21.3-2) = 80 * 4.18 * (21.3-22)

msteel = (80 * 4.18 * (-0.7)) / (0.452 * 19.3)

msteel = -234.08 / 8.7236

msteel = -26.833 g

Since mass can't be negative we should take the absolute value of it = 26.833g

The mass of the steel bar is 26.833 grams

6 0
3 years ago
What is the volume of 0.250 m hydrochloric acid required to react completely with 20.0 ml of 0.250 m ca(oh)2?
Soloha48 [4]
The balanced equation for the above neutralisation reaction is as follows;
Ca(OH)₂ + 2HCl ----> CaCl₂ + 2H₂O
Stoichiometry of Ca(OH)₂ to HCl is 1:2
number of Ca(OH)₂ moles reacted - 0.250 mol/L x 20.0 x 10⁻³ L = 5.00 x 10⁻³ mol
according to molar ratio of 1:2
number of HCl moles required = 2 x number of Ca(OH)₂ moles reacted
number of HCl moles = 5.00 x 10⁻³ x 2 = 10.0 x 10⁻³ mol
molarity of HCl solution - 0.250 M
there are 0.250 mol  in volume of 1 L
therefore 10.0 x 10⁻³ mol in - 10.0 x 10⁻³ mol  / 0.250 mol/L = 40.0 mL 
40.0 mL of 0.250 M HCl is required
5 0
3 years ago
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