Ask question

Ask question

All categories

Romashka-Z-Leto [24]

2 years ago

12

5. In a particular reaction 6.80g of dinitrogen trioxide gas (N203) was actually produc

1 answer:

puteri [66]2 years ago

4 0

Honestly just look at the question deeper and find it inside the equation of the fraction

You might be interested in

What mass of oxygen gas is consumed in a reaction that produces 4.60 mol sulfur dioxide?

babunello [35]

Answer:

jsjdhiefjcbjajrjrjwbxjuq ueei1udj

7 0

2 years ago

How can we remove the hardness of water by boiling?

pantera1 [17]

Heat makes it softer

5 0

3 years ago

How many grams of ca are needed to react completely with 2.20 L of a 4.50 m hcl solution

Dmitry_Shevchenko [17]

Ca + 2HCl = CaCl₂ + H₂

c=4.50 mol/l
v=2.20 l

n(HCl)=cv

m(Ca)/M(Ca)=n(HCl)/2

m(Ca)=M(Ca)cv/2

m(Ca)=40g/mol·4.50mol/l·2.20l/2=198 g

198 grams of Ca are needed

5 0

3 years ago

Read 2 more answers

Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.

Colt1911 [192]

Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

$\%C_E = \frac{E}{T} * 100\%$

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In $(Ca(OH)_2)$

The amount of all elements is: (i.e formula mass of $(Ca(OH)_2)$ )

$T = 1 * Ca + 2 * H + 2 * O$

$T = 1 * 40 + 2 * 1 + 2 * 16$

$T = 74$

The amount of calcium is: (i.e formula mass of calcium)

$E = 1 * Ca$

$E = 1 * 40$

$E = 40$

So, the percentage component of calcium is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{40}{74} * 100\%$

$\%C_E = \frac{4000}{74}\%$

$\%C_E = 54.05\%$

The amount of hydrogen is:

$E = 2 * H$

$E = 2 * 1$

$E = 2$

So, the percentage component of hydrogen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{2}{74} * 100\%$

$\%C_E = \frac{200}{74}\%$

$\%C_E = 2.70\%$

Similarly, for oxygen:

The amount of oxygen is:

$E = 2 * O$

$E = 2 * 16$

$E = 32$

So, the percentage component of oxygen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{32}{74} * 100\%$

$\%C_E = \frac{3200}{74}\%$

$\%C_E = 43.24\%$

5 0

2 years ago

A sample of gas occupies a volume of 445 mL at a temperature of 274 K. At what temperature will this sample of gas occupy a volu

Ksenya-84 [330]

V1 = 445ml V2 = 499ml
T1 = 274 K T2 = ?

By Charles Law,
V1/T1 = V2/T2
445/274 = 499/T2
By solving we get,
T2 = 307.25 K

4 0

2 years ago