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PtichkaEL [24]
3 years ago
7

Helpppppo meeeeee plzzzzzzzzz

Chemistry
1 answer:
antoniya [11.8K]3 years ago
6 0
For 2 it’s 2ft 10in
For 5 it’s 3ft 6in
You: put how tall you are
Adult: you can either measure an adult or use this 5 ft 10in
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¿Un kilo y medio a cuantos gramos equivale?
Anastaziya [24]

Answer:

Por tanto, podemos concluir que en medio kilogramo hay un total de 500 gramos

7 0
3 years ago
Suppose you have a dozen carbon atoms, a dozen gold atoms, and a dozen iron atoms. Even though you have the same number of each,
mixer [17]

Answer:

By weight they have the same mass, but the number of atoms is different

Explanation:

3 0
4 years ago
Read 2 more answers
Which process can separate out the solute from the solvent?
sineoko [7]
Evaporation of the solution
3 0
4 years ago
Please help me i would greatly appreciate it.. 50 points and will mark brainliest
Nikitich [7]

Answer:

368.92g

Explanation:

Firstly, let's balance the equation which is

2NO + O₂ ---> 2NO₂

Starting with 8.02 mol of NO let's calculate the moles of oxygen which is in a 2 : 1 molar ratio

2NO + O₂

2 : 1

8.02 mol : x mol

Moles of O₂ = 8.02 ÷ 2 = 4.01 mol

Doing the same thing for 18.75 mol of O₂ to calculate the number of moles of NO

2NO + O₂

2 : 1

x mol : 18.75 mol

Moles of NO = 18.75 × 2 = 37.5 however we are told we have 8.02 moles of NO, so we are unable to use 18.75 mol of O₂

Using 8.02 mol of NO to figure out the number of moles of NO₂ :

2NO : 2NO₂

They have the same molar ratio of 2 : 2, so the number of moles is 8.02

Using formula moles = mass / Molar mass

Rearranging to find mass = moles × molar mass

Molar mass of NO₂ = 14 + 16 + 16 = 46

Mass = 46 × 8.02 = 368.92g

5 0
2 years ago
Read 2 more answers
g compute the specific heat capacity at constant volume of nitrogen gas. the molar mass of N2 is 29.0 You warm 1.8 kg ov water a
Elis [28]

Complete question:

(a) compute the specific heat capacity at constant volume of nitrogen gas. the molar mass of N₂ is 29.0 You warm 1.8 kg of water at a constant volume 1.00 L from  21 C to 30.5 C in a kettle. For the same amount of heat, how many kilograms of 21∘C air would you be able to warm to 30.5∘C ?

(b) What volume (in liters) would this air occupy at 21∘C and a pressure of 1.00 atm? Make the simplifying assumption that air is 100% N₂​

Answer:

(a) The specific heat capacity of N₂ is 715.86 J/kg.K

(b) The volume the air occupy at 21∘C is 8784.29 Liters

Explanation:

Given;

M is the molar mass of N₂ = 29 x 10⁻³ kg/mol

specific heat of N₂ at constant volume, Cv = 20.76 J/mol.K

(a)

The specific heat capacity of N₂ is calculated as;

C = \frac{C_v}{M} \\\\C = \frac{20.76}{29 *10^{-3}} \\\\C = 715.86  \ J/kg.K

(b) heat capacity of water;

Q = mcΔθ

where;

c is the specific heat capacity of water = 4200 J/kg.K

m is mass of water, = 1.8 kg

Δθ is change in temperature, = 30.5 - 21 = 9.5 °C

Q = 1.8 x 4200 x  9.5

Q = 71820 J

Mass of nitrogen gas N₂, at this quantity of heat;

m_{N_2} = \frac{Q}{C*\delta \theta} \\\\m_{N_2} = \frac{71820}{715.86*9.5}\\\\m_{N_2} = 10.56 \ kg

The volume this air occupy at 21∘C

Apply ideal gas law;

PV = nRT = \frac{m}{M} RT

PV = \frac{mRT}{M} \\\\V = \frac{mRT}{MP}\\\\V = \frac{10.56(kg)*8.314*10^3(L.Pa/mol.K)*294(K)}{29*10^{-3}(kg)1.01325*10^5 (Pa)}\\\\V = 8784.29 \ Liters

6 0
4 years ago
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