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3 years ago

(WILL MARK BRAINLIEST) PLZ HELP ASAP

Chemistry

2 answers:

ArbitrLikvidat [17]3 years ago

3 0

Answer:

a) 2NaOH(aq) + CuSO4(aq) -------------> Cu(OH)2(s) + Na2SO4(aq)

b) Ca(OH)2(aq) + CO2(g) --------------> CaCO3 + H2O (this is already balanced)

c) Pb(NO3)2 + H2SO4 --------> PbSO4 + 2HNO3.

d) 2KNO3 ------> 2KNO2 + O2

e) H2SO4 + 2(NaOH) -----> Na2SO4 + 2(H2O)

f) Ca(NO3)2(aq) + (NH4)2CO3(aq) ----------------> CaCO3(s) + 2NH4NO3(aq)

ikadub [295]3 years ago

3 0

Answer:

a) 2NaOH(aq) + CuSO4(aq) ----> Cu(OH)2(s) + Na2SO4(aq)

b) Ca(OH)2(aq) + CO2(g) --> CaCO3 + H2O (the chemical equation is already balanced as is)

c) Pb(NO3)2 + H2SO4 --> PbSO4 + 2HNO3

d) 2KNO3 --> 2KNO2 + O2

e) H2SO4 + 2(NaOH) --> Na2SO4 + 2(H2O)

f) Ca(NO3)2(aq) + (NH4)2CO3(aq) --> CaCO3(s) + 2NH4NO3(aq)

Explanation:

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Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g)+ Br2

nasty-shy [4]

Answer:

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr ( g ) ↔ 2NO ( g ) + Br₂ ( g )

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,

Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2 [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ] ' ²

Substituting the new concentration terms ,

K ' = 1/2 [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ] ²

K ' = 1/4 [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ] ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' = [ NO ] ² [ Br₂ ] / [ NoBr ] ²

Hence ,

K' = K

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3 years ago

What is the percent yield if 107.50 g NH3 reacts with excess O2 according to the

Maksim231197 [3]

Answer:

81.59%

Explanation:

4NH₃ + 5O₂ → 4NO + 6H₂O

First we convert 107.50 g of NH₃ into moles, using its molar mass:

107.50 g NH₃ ÷ 17 g/mol = 6.32 mol NH₃

Now we calculate how many moles of NO would have been formed by the complete reaction of 6.32 moles of NH₃:

6.32 mol NH₃ * $\frac{4molNO}{4molNH_3}$ = 6.32 mol NO

Then we convert 6.32 moles of NO to grams, using its molar mass:

6.32 mol NO * 30 g/mol = 189.60 g NO

Finally we calculate the percent yield:

154.70 g / 189.60 g * 100% = 81.59%

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3 years ago

WILL GIVE BRANLIEST TO WHOEVER ANSWERS THIS QUESTION>>> How many valence electrons does NaHCo3 have?

nydimaria [60]

Answer:

NaHCO3

Na = 1

H = 1

C = 4

Explanation:

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To what temperature must a balloon, initially at 25°c and 2.00 l, be heated in order to have a volume of 6.00 l?

Kaylis [27]

V1 = 2.00 L
T1 = 25 + 273 = 298 K 
V2 = 6.00 L 
T2 = ? 
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What is the mass of the hydrogen atoms in 1 mole of water

Katen [24]

Answer:

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Explantion:

We are to find the mass of the hydrogen atoms in 1 mole of water.

We know that the formula of water is: $H_2O$