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3 years ago

Please help need 18-20 answer

Chemistry

2 answers:

CaHeK987 [17]3 years ago

8 0

18- So the equation for density is d=m/v. So if we take the volume of 500ml + 300ml we get a volume of 800ml and a mass of 400g. So we use the equation d=400g/800ml. This gives us a density of 1g/2ml.

19- if we have a ring with a mass of 107g and a volume of 5ml. We will use the following formula d=m/v so this ends up as 21.4=107g/5ml this gives us the material Platinum

Vlad [161]3 years ago

4 0

Answer:

800 ml

marbel

flow

Explanation:

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2. Consider the reaction 2NO(g) + O2(g) → 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is r

vredina [299]

Answer :

(a) The rate of $NO_2$ formed is, 0.066 M/s

(b) The rate of $O_2$ formed is, 0.033 M/s

Explanation : Given,

$\frac{d[NO]}{dt}$ = 0.066 M/s

The balanced chemical reaction is,

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate of disappearance of $NO$ = $-\frac{1}{2}\frac{d[NO]}{dt}$

The rate of disappearance of $O_2$ = $-\frac{d[O_2]}{dt}$

The rate of formation of $NO_2$ = $\frac{1}{2}\frac{d[NO_2]}{dt}$

As we know that,

$\frac{d[NO]}{dt}$ = 0.066 M/s

(a) Now we have to determine the rate of $NO_2$ formed.

$\frac{1}{2}\frac{d[NO_2]}{dt}=\frac{1}{2}\frac{d[NO]}{dt}$

$\frac{d[NO_2]}{dt}=\frac{d[NO]}{dt}=0.066M/s$

The rate of $NO_2$ formed is, 0.066 M/s

(b) Now we have to determine the rate of molecular oxygen reacting.

$-\frac{d[O_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}$

$\frac{d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s$

The rate of $O_2$ formed is, 0.033 M/s

6 0

3 years ago

Please help fast...please

RSB [31]

Answer:

Radioactive

Explanation:

hope this helped! :)

5 0

3 years ago

Chemistry vocabulary and answer

finlep [7]

Answer:

you forgot to attach a picture :)

Explanation:

8 0

3 years ago

The energy transfer between the system and the surroundings, in the form of heat at constant pressure is described by which of t

kicyunya [14]

I think the answer would be d but not 100% sure

4 0

4 years ago

2.70g of Zn (s) reacts with 50.0 mL of 1.00 M HCl solution to produce hydrogen gas according to the reaction. (R = 0.08206 L·atm

Anika [276]

Answer:

a. 0.0413 moles Zn

b. 0.0500 moles HCl

c. HCl is the limiting reactant

d. 0.0250 moles H₂

e. V = 0.56L

Explanation:

The reaction of Zn(s) with HCl is:

Zn(s) + 2HCl (aq) → ZnCl₂ (aq) + H₂(g)

Where 1 mole of Zn reacts with 2 moles of HCl.

a) To convert mass in grams to moles of a substance you need to use molar mass (Molar mass Zn: 65.38g/mol), thus:

2.70g Zn × (1mol / 65.38g) = 0.0413moles of Zn

b. Now, when you have a solution in molarity (Moles / L), you can know the moles of a volume of solution, thus:

Moles HCl:

50.0mL = 0.0500L × (1.00mol / L) = 0.0500 moles HCl

c. The limiting reactant is founded by using the chemical reaction as follows:

For a complete reaction of 0.0500 moles HCl you need:

0.0500 moles HCl × (1 mole Zn / 2 moles HCl) = 0.0250 moles Zn

As you have 0.0413 moles of Zn, and you need just 0.0250 moles for the complete reaction, Zn is the exces reactant and HCl is the limiting reactant

d.As HCl is limiting reactant and 2 moles of HCl react with 1 mole of H₂, moles of hydrogen formed are:

0.0500 moles HCl × (1 mole H₂ / 2 moles HCl) = 0.0250 moles H₂

e. Using PV = nRT, you can find volume of gas, thus:

PV = nRT

V = nRT / P

Where P is pressure 1atm at STP, n are moles, R is gas consant 0.08206Latm/molK and T is absolute temperature 273.15K at STP.

V = 0.0250molesₓ0.082atmL/molKₓ273.15K / 1atm

V = 0.56L

3 0

3 years ago

Please help need 18-20 answer ​

Please help need 18-20 answer