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2 years ago

Which indicator is blue in a solution that has a pH of 5.6?

Chemistry

2 answers:

m_a_m_a [10]2 years ago

7 0

Bromcresol green is the indicator that is blue in a solution that has a Ph of 5.6.

Vlad1618 [11]2 years ago

4 0

Answer:

1) bromcresol green

Explanation:

An acid base indicator is used to indicate the completion of an acid base titration experiment. Acid base indicators shows color change with a change in the pH of the solution. Bromcresol green, methyl orange, thymol blue, and bromthymol blue are acid base indicators. At pH of 5.6 bromcresol green is blue, methyl orange is yellow, thymol blue is yellow, and bromthymol blue is also yellow.

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Which example best illustrates a method to show differences among Earth’s layers?

hoa [83]

Answer:

Carson models how the continental crust varies in thickness. Marisol records atmospheric and oceanic temperatures at several beaches. Eliza analyzes seismic wave activity from an earthquake using a computer model.

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2 years ago

Read 2 more answers

5.00 g of hydrogen gas and 50.0g of oxygen gas are introduced into an otherwise empty 9.00L steel cylinder, and the hydrogen is

GenaCL600 [577]

1) Balanced chemical reaction:

2H2 + O2 -> 2H20

Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O

2) Reactant quantities converted to moles

H2: 5.00 g / 2 g/mol = 2.5 mol

O2: 50.0 g / 32 g/mol = 1.5625 mol

Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).

3) Products

H2 totally consumed -> 0 mol at the end

O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end

H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.

Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol

4) Pressure

Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K

p = nRT/V = 7.9 atm

3 0

2 years ago

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You find a little bit (0.150g) of a chemical marked Tri-Nitro-Toluene, and upon combusting it in oxygen, collect 0.204 g of CO2

avanturin [10]

1) Answer is: the formula is C₇N₃O₆H₅.

M(TNT) = 0.150 g; mass of the trinitrotoluene.

ω(N) = 18.5% ÷ 100%.

ω(N) = 0.185; mass percentage of the nitrogen.

m(N) = 0.150 g · 0.185.

m(N) = 0.02775 ·; mass of the nitrogen.

n(N) = 0.02775 g ÷ 14 g/mol.

n(N) = 0.002 mol; amount of the nitrogen.

n(CO₂) = 0.204 g ÷ 44 g/mol.

n(CO₂) = 0.0046 mol.

n(C) = n(CO₂) = 0.0046 mol; amount of the carbon.

m(C) = 0.0046 mol · 12 g/mol.

m(C) = 0.0552 g; mass of the carbon.

n(H₂O) = 0.030 g ÷ 18 g/mol.

n(H₂O) = 0.00166 mol.

n(H) = 2 · n(H₂O) = 0.0033 mol; amount of the hydrogen.

m(H) = 0.0033 mol · 1 g/mol.

m(H) = 0.0033 g; mass of the hydrogen.

2) m(O) = m(TNT) - m(N) - m(C) - m(H).

m(O) = 0.150 g - 0.02775 g - 0.0552 g - 0.0033 g.

m(O) = 0.06375 g.

n(O) = 0.06375 g ÷ 16 g/mol.

n(O) = 0.004 mol; amount of oxygen.

n(C) : n(N) : n(O) : n(H) = 0.0046 mol : 0.002 mol : 0.004 mol : 0.0033 mol.

n(C) : n(N) : n(O) : n(H) = 2.33 : 1 : 2 : 1.66 /×3.

n(C) : n(N) : n(O) : n(H) = 7 : 3 : 6 : 5.

8 0

2 years ago

Beer brewing begins with steeping grains in hot water, releasing the sugars inside. The sugar water is then heated to a boil and

user100 [1]

Answer:

The answers to the question are

a. 166.64 ° F

b. 217990.08 J/hour or 60.55 J/s = 60.55 watts

c. 13.C

Explanation:

a. To solve the question we list out the given variables thus

mass of grain = 16.5 lbs

Temperature of grain = 67 °F

Volume of hot water = 5 gals = ‪0.02273‬ m³

Equilibrium temperature of the mixture = 154 °F

Specific heat capacity of the grain = 0.44 times specific heat capacity of water

Therefore we have

Heat supplied by hot water = heat gained by mixture

Density of the water = 997 kg/m³ which gives

Therefore the mass of the water = (Density of the water) × (Volume of the water) = (997 kg/m³) × ‪(0.02273‬ m³) = 22.66181 kg

Therefore the heat supplied by the water =22.66 kg×1000 g/kg ×4.2 J/g°C×(Tₓ -‪67.78 °C) = ‪7.48 kg×1000 g/kg×0.44×4.2 J/g°C×(67.78 -‪19.44)

= 95172 × (Tₓ -‪67.78 °C) =668205.7536 J

(Tₓ -‪67.78 °C) = 7.02 from where Tₓ = 74.80 °C = ‪166.64 ° F

The initial temperature (strike temperature) of the hot water = 74.80 °C = 166.64 ° F

b. Where the mixture lost two degrees we have

22.66 kg×1000 g/kg ×4.2 J/g°C×2 °C + ‪7.48 kg×1000 g/kg×0.44×4.2 J/g°C×2 °C = 217990.08 J therefore the average energy lost per unit time = 217990.08 J/hour or 60.55 J/s

c. To find out how much it cost we have

Heat energy required to raise 5 gallons of water from 110 °F to 166.64 °F we have

22.66 kg×1000 g/kg ×4.2 J/g°C×(74.8 °C-‪43.33 °C) = 2994745.92 J

Energy lost during the heating = 10% = 299474.59 J

Total energy supplied 2994745.92 J + 299474.59 J = 3294220.5 J

Time for heating = 47 minutes, therefore rate of energy consumption = (3294220.5 J)/ (47×60) = 1168.163 Watt 1.168 kW

Cost of energy = 15.C per kilowatt-hour therefore 1.168 kW for 47 minutes will cost

1.168 kW ×47/60×15 = 13.C

therefore it cost 13.C to heat the 5 gallons of tap water initially at 110 ° F to the strike temperature 166.64 °F

6 0

2 years ago

Red blood cells are placed in a solution and neither hemolysis nor crenation occurs. therefore the solution is

Len [333]

The answer is isotonic solution. These are solutions where the solute concentration in the solution and inside the cells are levelled and consequently water flows consistently. When red blood cells are positioned in an isotonic solution the cells would always stay the same.

3 0

2 years ago