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Nonamiya [84]
3 years ago
7

The complete combustion of C4H10, will require how many moles of oxygen, O2? ________.​

Chemistry
1 answer:
tangare [24]3 years ago
8 0

Answer:

2

Explanation:

C⁴H¹⁰ + 2O² —> 4CH²O + H²

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A balanced chemical equation used to prepare ammonium carbonate, (nh4)2co3 , is: 2 n h 3 ( g ) + c o 2 ( g ) + h 2 o ( l ) ⟶ ( n
Tems11 [23]

Answer:

This a simple stoichiometry problem using the ideal gas law.  

First take the grams of ammonium carbonate and convert it to moles using its molar mass and dividing. 11.9 g/96.0932 g/mol= .12384 mol  

Now use a molar conversion using the balanced equation,  

1 mol (NH4)2CO3 ---> 4 mol gas formed (2 mol NH3 + 1 mol CO2 + 1 mol H2O) = .12384 x 4 = .49535 mol gas  

PV=nRT  

V=nRT/P= .49535mol (.08206 Lxatm/molxK) (296K)/ (1.03 atm)=11.682 L

7 0
3 years ago
What is the name for s2o7
Arada [10]

Answer:

Disulfate Ion {2-} Name: Disulfate Ion {2-} Formula: S2O7.

Explanation:

HOPE THIS HELPED :)

6 0
3 years ago
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Chloroform has a density of 1.483g/ml. A sample of chloroform has a volume of 1.93 liters. How many kilograms does it weight? ​
ollegr [7]

Weight of Chloroform : = 2.862 kg

<h3>Further explanation</h3>

Given

Density 1.483 g/ml

Volume = 1.93 L

Required

Weight of Chloroform

Solution

Density is a quantity derived from the mass and volume  

Density is the ratio of mass per unit volume  

Density formula:  

\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}

ρ = density  

m = mass  

v = volume  

Convert density to kg/L :

=1.483g/ml = 1.483 kg/L

So the weight(mass) :

= ρ x V

= 1.483 kg/L x 1.93 L

= 2.862 kg

8 0
3 years ago
A 500.0-mL buffer solution is 0.100 M in HNO2 and 0.150 M in KNO2. Determine whether each addition would exceed the capacity of
Leviafan [203]

Answer:

None of the additions will exceed the capacity of the buffer.

Explanation:

As we know a buffer has the ability to resist pH changes when small amounts of strong acid or base are added.

The pH of the buffer is given by the Henderson-Hasselbach equation:

pH = pKa + log [A⁻] / [HA]

where A⁻ is the conjugate base of the weak acid HA.

Now we can see that what is important is the ratio [A⁻] / [HA] to resist a pH change brought about by the addition of acid or base.

It follows then that once we have consumed by neutralization reaction either the acid or conjugate base in the buffer, this will lose its ability to act as such and the pH will increase or decrease dramatically by any added acid or base.

Therefore to solve this question we must determine the number of moles of acid HNO₂ and NO₂⁻ we have in the buffer and compare it with the added acid or base to see if it will deplete one of these species.

Volume buffer = 500.0 mL = 0.5 L

# mol HNO₂ = 0.5 L x 0.100 mol/L = 0.05 mol HNO₂

# mol NO₂⁻ = 0.5 L x 0.150 mol/L = 0.075 mol NO₂⁻

a. If we add 250 mg NaOH (0.250 g)

molar mass NaOH =40 g/mol

# mol NaOH =0.250 g/ 40g/mol = 0.0063 mol

0.0063 mol NaOH will be neutralized by 0.0063 mol HNO₂ and we have plenty of it, so it would not exceed the capacity of the buffer.

b. If we add 350 mg KOH (0.350 g)

molar mass KOH =56.10 g

# mol KOH = 0.350 g/56.10 g/mol = 0.0062 mol

Again the capacity of the buffer will not be exceeded since we have 0.05 mol HNO₂ in the buffer.

c. If we add 1.25 g HBr

molar mass HBr = 80.91 g/mol

# mol HBr = 1.25 g / 80.91 g/mol = 0.015 mol

0.015 mol Hbr will neutralize 0.015 mol NO₂⁻ and we have to start with 0.075 mol in the buffer, therefore the capacity will not be exceeded.

d. If we add 1.35 g HI

molar mass HI = 127.91 g/mol

# mol HI = 1.35 g / 127.91 g/mol = 0.011 mol

Again the capacity of the buffer will not be exceed since we have plenty of it in the buffer after the neutralization reaction.

7 0
3 years ago
A theater director reinterprets a play by replacing its classical score with
cricket20 [7]

Answer:

The play will be more  appealing to a younger audience.

Explanation:

A younger audience will more likely appreciate current pop hits rather than classical score.

4 0
3 years ago
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