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2 years ago

13

What is the shape of the waterline?

1 answer:

Vikentia [17]2 years ago

5 0

The waterline is where the hull of a ship meets the surface of the water, it is also a special marking indicating the legal load limit of a ship, and, in naval architectural plans, it is any line drawn to delineate hull shape that is in a plane parallel to the surface of the water.

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Round off each number to the indicated number of significant figures (sf)

weeeeeb [17]

Answer:

The answer is B

Explanation:

3 0

2 years ago

Read 2 more answers

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Mariana [72]

9.Pubchem
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2 years ago

Fe3+(aq) (yellow) + SCN-(aq) (colorless) FeSCN2+(aq) (blood-red) Chloride ions are colorless. Potassium ions are also colorless.

harina [27]

Explanation:

$Fe^{3+}\text{(aq)(yellow)}+SCN^-\text{(aq)(colorless)}\rightleftharpoons FeSCN^{2+}\text{(aq)(blood-red)}$

$K^+\text{(aq)(colorless)}+Cl^-\text{(aq)(colorless)}\rightarrow KCl$

The above two reactions xcan also be written in form of single chemcial equation:

$FeCl_3+K(SCN)\rightleftharpoons KCl+[Fe(SCN)]Cl_2(blood-red)$

1. Color of ferric chloride solution is yellow. This is due to presence of ferric ions which have yellow color in their aqueous solutions.

2. KSCN has the colorless solution. This due to potassium ion forms colorless aqueous solution.

3. On mixing, KSCN with $FeCl_3$ we will get blood red color solution of $[FeSCN]Cl_2$ .

5 0

3 years ago

Calculate the heat absorbed when 31.6 g of ice at 0°C is completely melted

attashe74 [19]

Latent heat of fusion for ice is 33600J/k

Mass=31.6J/k

Heat absorbed

k_f×m
31.6×33600
1061,740mJ
1061J

8 0

2 years ago

Wastewater from a cement factory contains 0.280 g of Ca2+ ion and 0.0220 g of Mg2+ ion per 100.0 L of solution. The solution den

faltersainse [42]

<u>Answer:</u> The concentration of $Ca^{2+}\text{ and }Mg^{2+}$ ions are 2.797 ppm and 0.212 ppm respectively.

<u>Explanation:</u>

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Volume of gold = 100 L = 100000 mL (Conversion factor: 1 L = 1000 mL)

Density of gold = 1.001 g/mL

Putting values in above equation, we get:

$1.001g/mL=\frac{\text{Mass of solution}}{100000mL}\\\\\text{Mass of solution}=1.001\times 10^5g$

To calculate the concentration in ppm (by mass), we use the equation:

$ppm=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6$

<u>Calculating the concentration of calcium ions:</u>

Mass of $Ca^{2+$ ions = 0.280 g

Putting values in above equation, we get:

$ppm(Ca^{2+})=\frac{0.280g}{1.001\times 10^5}\times 10^6=2.797ppm$

<u>Calculating the concentration of magnesium ions:</u>

Mass of $Mg^{2+$ ions = 0.0220 g

Putting values in above equation, we get:

$ppm(Mg^{2+})=\frac{0.0220g}{1.001\times 10^5}\times 10^6=0.212ppm$

Hence, the concentration of $Ca^{2+}\text{ and }Mg^{2+}$ ions are 2.797 ppm and 0.212 ppm respectively.

7 0

3 years ago