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faltersainse [42]
3 years ago
13

Name the gas that the limewater in the flask is testing for.

Chemistry
1 answer:
enyata [817]3 years ago
4 0

Answer:

Explanation:

Carbon dioxide is the answer I think.

Hope it helps...

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II need this asap !!!!!!!
Xelga [282]

Answer:

1. B.10 mol/l

2. D.

3.A.

Explanation:

Totoo po yan ganyan po yong sakin

7 0
2 years ago
Cane sugar is being manufactured at the rate of 500 kg/hr from sugar canes that have the following composition (% by weight):
My name is Ann [436]

Answer:

a. 4166,67 b. 74553.0

Explanation:

6 0
3 years ago
The combustion of gasoline produces carbon dioxide and water. Assume gasoline to be pure octane (C8H18) and calculate the mass (
Mariulka [41]

Answer:

3.09kg

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

2C8H18 + 25O2 —> 16CO2 + 18H2O

Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol

Mass of C8H18 from the balanced equation = 2 x 114 = 228g

Converting 228g of C8H18 to kg, we obtained:

228/1000 = 0.228kg

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 16 x 44 = 704g

Converting 704g of CO2 to kg, we obtained:

704/1000 = 0.704kg

From the equation,

0.228kg of C8H18 produced 0.704kg of CO2.

Therefore, 1kg of C8H18 will produce = 0.704/0.228 = 3.09kg of CO2

6 0
3 years ago
Reacting 35.4 ml of 0.220 m agno3 with 52.0 ml of 0.420 m k2cro4 results in what mass of solid formed
laila [671]
Answer is: 1.29 grams <span>of solid formed.
</span>Chemical reaction: 2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq).
n(AgNO₃) = c(AgNO₃) · V(AgNO₃).
n(AgNO₃) = 0.220 M · 0.0351 L.
n(AgNO₃) = 0.0078 mol; limiting reactant.
n(K₂CrO₄) = 0.420 M · 0.052 L.
n(K₂CrO₄) = 0.022 mol.
From chemical reaction: n(AgNO₃) : n(Ag₂CrO₄) = 2 : 1.
n(Ag₂CrO₄) = 0.0078 mol ÷ 2.
n(Ag₂CrO₄) = 0.0039 mol.
m(Ag₂CrO₄) = 0.0039 mol · 331.73 g/mol.
m(Ag₂CrO₄) = 1.29 g.

7 0
3 years ago
Enter your answer in the provided box. Liquid methanol (CH3OH) can be used as an alternative fuel in pickup and SUV engines. An
andreyandreev [35.5K]

Answer: E=∆H*n= -40.6kj

Explanation:

V(CO) =15L=0.015M³

P=11200Pa

T=85C=358.15K

PV=nRT

n=(112000×0.015)/(8.314×358.15)

n(Co)= 0.564mol

V(Co)= 18.5L = 0.0185m³

P=744torr=98191.84Pa

T= 75C = 388.15k

PV=nRT

n= (99191.84×0.0185)/(8.314×348.15)

n(H2) = 0.634mol

n(CH30H) =1/2n(H2)=1/2×0.634mol

=0.317mol

∆H =∆Hf{CH3OH}-∆Hf(Co)

∆H=-238.6-(-110.5)

∆H = 128.1kj

E=∆H×n=-40.6kj.

3 0
3 years ago
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