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inysia [295]
3 years ago
5

Helpppppppppppp me pleaseeeeeee​

Chemistry
1 answer:
Cloud [144]3 years ago
4 0

Answer:

VX a small town called to the highest frequency and I wla kmo the

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41 to
tia_tia [17]

Answer:

d = 0.9 g/L

Explanation:

Given data:

Number of moles = 1 mol

Volume = 24.2 L

Temperature = 298 K

Pressure = 101.3 Kpa (101.3/101 = 1 atm)

Density of sample = ?

Solution:

PV = nRT     (1)

n = number of moles

number of moles = mass/molar mass

n = m/M

Now we will put the n= m/M in equation 1.

PV = m/M RT   (2)

d = m/v

PM = m/v RT ( by rearranging the equation 2)

PM = dRT

d = PM/RT

The molar mass of neon is = 20.1798 g/mol

d = 1 atm × 20.1798 g/mol / 0.0821 atm. L/mol.K × 273K

d = 20.1798 g/22.413 L

d = 0.9 g/L

4 0
3 years ago
PLEASE ANSWER CORRECTLY FOLLOWING GUIDLINES SO ANSWER WONT GET DELETED I REALLY NEED HELP
blagie [28]

Chemical reaction

equation

reactants

products

yields

mass

balanced

atoms

coefficients

numbers

element

correct

substance

two

Explanation:

did my best lol I'm like 98.69% confident

8 0
2 years ago
How many liters of hydrogen are needed to produce 34 grams NH3
polet [3.4K]
You would need 1000 liters
8 0
3 years ago
A student with an unknown code of Q3A conducts a qualitative analysis of her unknown. Her results indicated the formation of a y
insens350 [35]

Answer:

Grey precipitate implies the presence of silver ions

Yellow precipitate implies the presence of lead II ions

Explanation:

Qualitative analysis provides us a quick method of identifying ions present in a sample by chemical reactions involving simple reagents. Precipitates having a unique colour is formed. The identity of ions in the sample is deduced from the colour of precipitate obtained when particular reagents are added.

In the question, a precipitate containing silver ions upon standing turn into grey colour. Similarly, lead II ions give a yellow precipitate.

6 0
3 years ago
A 31.1 g wafer of pure gold, initially at 69.3 _c, is submerged into 64.2 g of water at 27.8 _c in an insulated container. what
KIM [24]
Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water 

Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.

From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water

At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
      = (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)- 
      = 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
       = (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
       = 268.356(T - 27.8)

Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C

Answer: 28.4 °C

3 0
3 years ago
Read 2 more answers
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