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riadik2000 [5.3K]
3 years ago
14

Is there any compound like O2C NO2​

Chemistry
1 answer:
SIZIF [17.4K]3 years ago
5 0

Answer:

Aye Wassup? Adam Here :D

Explanation:

Nitrogen dioxide is a chemical compound with the formula NO 2.It is one of several nitrogen oxides. NO 2 is an intermediate in the industrial synthesis of nitric acid, millions of tons of which are produced each year for use primarily in the production of fertilizers.At higher temperatures it is a reddish-brown gas.

Hope This Helped Bro: "AdamDaAssasin"

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What does the coefficients in a balanced chemical equation represent​
attashe74 [19]

Answer:

th mol

Explanation:

When you balance an equation, you can refer back to the coefficients when calcuating mol and molar mass from the given information about one reactant to antoher reactant ot product. you use the coeffiecients for mol-mol ratio.

for example. if you are given A +3 B -----> 6C + 5D, and you have 10 mol of A and you have to find how many mols of C with 10 mol of A. then you would ratio it as A: C = 10A: 10C= 10(1): 10(6)= 10:60. so with 10 mol A you can make 60 mol C.

8 0
3 years ago
Calculate the ph of the solution 0.2M Mg(OH)2
Kipish [7]

Mg(OH)2 is base

Oh = 0.2 x 2 = .4

Poh = - Log (.4)

= .3979

Ph + Poh = 14

Ph = 14 - .3979

= 13.60

6 0
3 years ago
Read 2 more answers
Read the statement.
lina2011 [118]

Answer:

60 g/L is the final concentration of NaI solution .

Explanation:

Molarity of NaI solution before evaporation =M_1= 0.2 M

Volume of NaI solution before evaporation =V_1= 2.0 L

Molarity of NaI solution after evaporation =M_2= ?

Volume of NaI solution after evaporation =V_2= 1.0 L

M_1V_1=M_2V_2 ( dilution)

M_2=\frac{M_1V_1}{V_2}=\frac{0.2 M\times 2.0 L}{1.0 L}=0.4 M

Molar mass of NaI = 150 g/mol

Concentration of NaI after evaporation :

0.4 M × 150 g/mol = 60 g/L

60 g/L is the final concentration of NaI solution .

4 0
3 years ago
The scientific theory of plate tectonics states that the Earth's crust is composed of many plates, or slabs of rock. Interaction
jonny [76]
The answer would be d
4 0
3 years ago
14. In the lab, an experimenter mixes 75.0 g of water (initially at 30oC) with 83.8 g of a solid metal (initially at 600oC). At
avanturin [10]

Answer:

Tungsten is used for this experiment

Explanation:

This is a Thermal - equilibrium situation. we can use the equation.

Loss of Heat of the Metal = Gain of Heat by the Water

                      -Q_{m}=+Q_{w}\\

                    Q = mΔTC_{p}

Q = heat

m = mass

ΔT = T₂ - T₁

T₂ = final temperature

T₁ = Initial temperature

Cp = Specific heat capacity

<u>Metal</u>

m = 83.8 g

T₂ = 50⁰C

T₁ = 600⁰C

Cp = x

<u>Water</u>

m = 75 g

T₂ = 50⁰C

T₁ = 30⁰C

Cp = 4.184 j.g⁻¹.⁰c⁻¹

               -Q_{m}=+Q_{w}\\

⇒ - 83.8 x x x (50 - 600) = 75 x 4.184 x (50 - 30)

⇒ x = \frac{6276}{46090} = 0.13 j.g⁻¹.⁰c⁻¹

We know specific heat capacity of Tungsten = 0.134 j.g⁻¹.⁰c⁻¹

So metal Tungsten used in this experiment

8 0
4 years ago
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