SOMEONE HELP ME ASAP!!!

Five beans have a mass of 2.1 g. How many beans are in 0.454 kg go beans?

Svetach [21]

<span>1080.9523809 should be the answer</span>

4 0

3 years ago

You want to prepare 500.0 mL of 1.000 M KNO3 at 20°C, but the lab (and water) temperature is 24°C at the time of preparation. Ho

timama [110]

Explanation:

As per the given data, at a higher temperature, at $24^{o}C$ , the solution will occupy a larger volume than at $20^{o}C$ .

Since, density is mass divided by volume and it will decrease at higher temperature.

Also, concentration is number of moles divided by volume and it decreases at higher temperature.

At $20^{o}C$ , density of water=0.9982071 g/ml

Therefore, $\frac{concentration}{density}$ will be calculated as follows.

= $\frac{C_{1}}{d_{1}}$

= $\frac{1.000 mol/L}{0.9982071 g/ml}$

= 1.0017961 mol/g

At $24^{o}C$ , density of water = 0.9972995 g/ml

Since, $\frac{concentration}{density}$ = $\frac{C_{2}}{d_{2}}$

= $\frac{C_{2}}{0.9972995}$

Also, $\frac{C_{1}}{d_{1}}$ = $\frac{C_{2}}{d_{2}}$

so, 1.0017961 mol/g = $\frac{C_{2}}{0.9972995}$

$C_{2} = 1.0017961 \times 0.9972995$

= 0.9990907 mol/L

Therefore, in 500 ml, concentration of $KNO_{3}$ present is calculated as follows.

$C_{2}$ = $\frac{concentration}{volume}$

0.9990907 mol/L = $\frac{concentration}{0.5 L}$

concentration = 0.49954537 mol

Hence, mass (m'') = $0.49954537 mol \times 101.1032 g/mol$ = 50.5056 g (as molar mass of $KNO_{3}$ = 101.1032 g/mol).

Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.

Hence, using Buoyancy correction as follows,

m = $m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}$

where, $d_{air}$ = density of air = 0.0012 g/ml

$d_{weight}$ = density of callibration weights = 8.0g/ml

d = density of weighed object

Hence, the true mass will be calculated as follows.

True mass(m) = $50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}$

true mass(m) = 50.5268 g

= 50.53 g (approx)

Thus, we can conclude that 50.53 g apparent mass of $KNO_{3}$ needs to be measured.

8 0

3 years ago

The forces between water molecules are stronger than the forces between ethanol molecules. Which liquid would probably be most d

xxTIMURxx [149]

An insect would have a harder time walking on ethanol than it would water. Water has stronger intermolecular forces which leads to higher surface tension. The higher surface tension allows for the insect to be easily supported

8 0

3 years ago

Read 2 more answers

Identify the name of the ionic compound with the formula MgS. Group of answer choices magnesium sulfide magnesium sulfite magnes

klasskru [66]

<h2>Answer: magnesium sulfide</h2>

<h3>Explanation:</h3>

MgS → Mg²⁺ + S²⁻

S²⁻ is known as the sulfide ion. That makes MgS Magnesium Sulfide.

For consideration:

magnesium sulfide MgS

magnesium sulfite MgSO₃

magnesium sulfate MgSO₄

magnesium sulfur Mg & S

6 0

3 years ago

You wish to calculate the mass of hydrogen gas that can be prepared from 5.06 g of srh2 and 4.34 g of h2o.

Alexus [3.1K]

<span>0.228 g The balance formula for the reaction between SrH2 and H2O is SrH2(s) + 2 H2O(l) ==> Sr(OH)2(s) + 2 H2(g) So for every mole of SrH2 used, 2 moles of hydrogen gas, or 4 moles of hydrogen atoms are released. So let's calculate the molar mass of SrH2 and H2O so see what the limiting reactant is. strontium = 87.62 Hydrogen = 1.00794 Oxygen = 15.999 Molar mass of SrH2 = 87.62 + 2 * 1.00794 = 89.63588 g/mol Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488 g/mol Moles of SrH2 = 5.06 g / 89.63588 g/mol = 0.056450609 mol Moles of H2O = 4.34 g / 18.01488 g/mol = 0.240911957 mol Looking at the balanced formula, for every mole of SrH2, it takes 2 moles of H2O. So the limiting reactant will be the SrH2. And for every mole of SrH2 used, we get 4 moles of hydrogen atoms. So 4 * 0.056450609 mole * 1.00794 g/mole = 0.227595307 g Since we only have 3 significant figures, round the result to 3 figures, giving 0.228 g</span>

5 0

4 years ago