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dsp73
3 years ago
10

Plz answer this fast... will mark ur answer as brainliest

Chemistry
2 answers:
Xelga [282]3 years ago
7 0

Answer:

When fuels are incompletely burnt, they release carbon monoxide gas into the atmosphere. This gas is very dangerous as it is poisonous in nature

Explanation:

Nataly [62]3 years ago
7 0

Answer:

:they release carbon monoxide gas into the atmosphere. This gas is very dangerous as it is poisonous in nature

Explanation

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Sacrificial protection takes place when a magnesium block is
Liula [17]

Answer:

Explained below.

Explanation:

In oil rigs, Sacrificial protection is the protection of the steel against corrosion by using magnesium block which is a more reactive metal. This means that steel which is to be protected from corrosion will be the cathode while magnesium block preventing it from corrosion will be the anode and as such the block will be the one to corrode instead.

The mechanism behind this sacrificial protection is that the steel pipe on the rig will be connected to the magnesium block through copper wires and then the magnesium block will donate its electrons to the steel thereby preventing it from rusting. Thus is a reversible reaction and as such, after the steel Iron undergoes oxidization by receiving electrons from the magnesium block, it will be immediately be reduced to iron as depicted in the equation below.

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3 years ago
A solution contains 482.6 g
mariarad [96]

Answer:

\large \boxed{\text{102.57 $\, ^{\circ}$C}}

Explanation:

Data:

m₂ = 482.6 g sucrose

m₁ = 0.2804 kg water

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{\text{b}} = iK_{\text{b}}b

i is the van’t Hoff factor —  the number of moles of particles you get from 1 mol of solute. For sucrose, i = 1.

1. Moles of sucrose

n = \text{482.6 g}  \times \dfrac{\text{1 mol}}{\text{342.30 g}} = \text{1.410 mol}

2. Molal concentration of  sucrose

b =  \dfrac{\text{1.410 mol}}{\text{0.2804 kg}} = \text{5.029 mol/kg}

3. Increase in boiling point

\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 5.029 \cdot mol \cdot kg^{-1} = 2.574 \, ^{\circ}\text{C}

4. Boiling point

\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 2.574 \, ^{\circ}\text{C} = \mathbf{102.57 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{102.57 \, ^{\circ}C}}$}

 

7 0
3 years ago
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