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3 years ago

For ICP. Please help

Chemistry

1 answer:

Schach [20]3 years ago

4 0

Answer:

its letter B.

Explanation:

If an avalanche occurs, the snow on the mountain accelerates down slope, converting more gravitational potential energy to kinetic energy.

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What is the mass of 2.5 mol of ca, which has a molar mass of g/mol

swat32

Atomic mass Ca = 40 a.m.u

1 mole Ca ----------- 40 g
2.5 mols Ca -------- ( mass Ca )

Mass Ca = 2.5 x 40 / 1

Mass Ca = 100 / 1

= 100 g of Ca

hope this helps!

3 0

3 years ago

In the laboratory you are asked to make a 0.565 m sodium bromide solution using 315 grams of water. How many grams of sodium bro

Scorpion4ik [409]

Answer : The mass of sodium bromide added should be, 18.3 grams.

Explanation :

Molality : It is defined as the number of moles of solute present in kilograms of solvent.

Formula used :

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}$

Solute is, NaBr and solvent is, water.

Given:

Molality of NaBr = 0.565 mol/kg

Molar mass of NaBr = 103 g/mole

Mass of water = 315 g

Now put all the given values in the above formula, we get:

$0.565mol/kg=\frac{\text{Mass of NaBr}\times 1000}{103g/mole\times 315g}$

$\text{Mass of NaBr}=18.3g$

Thus, the mass of sodium bromide added should be, 18.3 grams.

6 0

3 years ago

A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

Answer: The percent yield of the 1-bromobutane is 48.65 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NaBr:

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0

3 years ago

How can we regain the hardness of water by adding lime?

Maru [420]

Answer:

due to the bicarbonate of CaCO3

5 0

3 years ago

Read 2 more answers

Lastly, Snape thinks we should try one more calculation. What is the retention factor if the distance traveled by the solvent fr

Neporo4naja [7]

Answer:

0.1 is the retention factor.

Explanation:

Distance covered by solvent , $d_s= 2.0 cm$

Distance covered by solute or ion, $d = 0.20 cm$

Retention factor $(R_f)$ is defined as ratio of distance traveled by solute to the distance traveled by solvent.

$R_f=\frac{d}{d_s}$

$R_f=\frac{0.20 cm}{2.0 cm}=0.1$

0.1 is the retention factor.

7 0

3 years ago