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3 years ago

State the difference between sugar in water and pebbles in water

1 answer:

6 0

Sugar is made of molecules that are bonded together based on the positively and negatively charged areas. They will slowly dissolve in water. Pebbles are solids. They will sit in water for a long time. Though shale pebbles will break apart or fall apart.

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A wave at sea travel with a velocity of 25 m/s. If it has a wavelength of 10 m, what is it’s frequency?

stealth61 [152]

Answer:

2.5 Hz

Explanation:

Formula for the given question is:

Velocity= frequency X Wavelength.

v= f X λ.

25 = 10 X f

f= 2.5 Hz

4 0

3 years ago

The outermost electrons are the_____electrons

Lynna [10]

Answer:

Valence electrons. they can be used to determine the group of an element.

Explanation:

6 0

3 years ago

The reaction 2HgO (s)→2Hg (I)+O2 (g) has a percent yield of 50%. You want to produce 100 g of Hg.

nevsk [136]

The mass of HgO needed for the reaction is 216 g

The correct answer to the question is Option C. 216 g

We'll begin by calculating the theoretical yield of Hg.

Actual yield of Hg = 100 g

Percentage yield = 50%

Theoretical yield of Hg =?

Theoretical yield = Actual yield / percentage yield

Theoretical yield = 100 / 50%

Theoretical yield of Hg = 200 g

Finally, we shall determine the mass of HgO needed for the reaction.

2HgO → 2Hg + O₂

Molar mass of HgO = 201 + 16 = 217 g/mol

Mass of HgO from the balanced equation = 2 × 217 = 434 g

Molar mass of Hg = 201 g/mol

Mass of Hg from the balanced equation = 2 × 201 = 402 g

From the balanced equation above,

402 g of Hg were produced from 434 g of HgO.

Therefore

200 g of Hg will be produce by = (200 × 434) / 402 = 216 g of HgO.

Thus, 216 g of HgO is needed for the reaction.

Learn more about stoichiometry:

brainly.com/question/24426334

4 0

2 years ago

Read 2 more answers

1. A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0%

bagirrra123 [75]

Answer:

$m_{PbI_2}=18.2gKI$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2KI+Pb(NO_3)_2\rightarrow 2KNO_3+PbI_2$

Thus, we proceed to compute the reacting moles of Pb(NO3)2 and KI, by using the given concentrations and densities and molar masses which are 331.2 g/mol and 166 g/mol respectively:

$n_{Pb(NO_3)_2}=96.7mL*\frac{1.134g}{mL}*\frac{0.14gPb(NO_3)_2}{1g}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} =0.0464molPb(NO_3)_2\\\\n_{KI}=99.8mL*\frac{1.093g}{mL}*\frac{0.12gKI}{1g}*\frac{1molKI}{166gKI} =0.0789molKI$

Next, the 0.0464 moles of Pb(NO3)2 will consume the following moles of KI (consider their 1:2 molar ratio):

$n_{KI}^{consumed\ by\ Pb(NO_3)_2}=0.0464molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0928molKI$

Hence, as only 0.0789 moles of KI are available, KI is the limiting reactant, therefore the formed grams of PbI2, considering its molar mass of 461.01 g/mol and 2:1 molar ratio, are:

$m_{PbI_2}=0.0789molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gKI$

Best regards.

3 0

4 years ago

what is the boiling point of water at the top of mount everest where the atmospheric pressure is only 34% as strong

Tema [17]

This question is incomplete, the complete is;

Water has a heat of vaporization (ΔHvap) of 44.01 kJ mol-1 and boils at 100 degrees C at sea level.

What is the boiling point of water at the top of Mount Everest where the atmospheric pressure is only 34% as strong as the pressure at sea level?

Answer: the boiling point of water at the top of mount Everest where the atmospheric pressure is only 34% as strong is 74 °C

Explanation:

Given that;

P1 = 1 atm,

T1 = 100°C = 373 K

P2 = P1 × 34% = 1×0.34 = 0.34 atm

T1 = ?

ΔH = 44.01 kJ mol⁻¹ = 44.01×10³ J/mol

R = 8.314 J/k.mol

now using the Clausius - Clapeyron equation;

p1_∫^p2 d.InP = ΔHvap/R T1_∫^T2. 1/T².dT

⇒ In(P2/P1) = ΔHvap/R (1/T1 - 1/T2)

so we substitute;

ln( 0.34/1 ) = ( 44.01×10³ / 8.314) × ( 1/373 - 1/T2)

-1.0788 = 5293.4808 × ( 0.00268 - 1/T2)

-1.0788 = 14.1865 - 5293.4808(1/T2)

5293.4808(1/T2) = 14.1865 + 1.0788

5293.4808(1/T2) = 15.2653

(1/T2) = 15.2653 / 5293.4808

(1/T2) = 0.0028837

T2 = 1 / 0.0028837

T2 = 346.8 K

WE convert to Celsius

t2 =346.8 K − 273.15 = 73.65 °C ≈ 74 °C

Therefore, the boiling point of water at the top of mount Everest where the atmospheric pressure is only 34% as strong is 74 °C

8 0

3 years ago