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3 years ago

Which of these metals would you expect to be found in nature as a deposit of relatively pure element?

1 answer:

7 0

1. Which of these metals would you expect to be found in nature as a deposit of relatively pure element? The Answer is: Lithium. And question 2 is: Potassium - furthest away from lead in the reactivity series.

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Which formula equation represents the burning of sulfur to produce sulfur dioxide?

Leni [432]

Answer:

S(s) + O2(g) --> SO2(g)

Upper S (s) plus upper O subscript 2 (g) right arrow with delta above upper S upper O subscript 2 (g).

Explanation:

The reaction is given as;

Sulfur + oxygen --> Sulphur dioxide

Sulphur = S

Oxygen = O2

Sulfur dioxide = SO2

So we have;

S(s) + O2(g) --> SO2(g)

The crrect option is option A. Upper S (s) plus upper O subscript 2 (g) right arrow with delta above upper S upper O subscript 2 (g).

5 0

3 years ago

Read 2 more answers

How many grams of nano3 would you add to 500g of h2o in order to prepare a solution that is 0.500 molal in nano3?

VARVARA [1.3K]

When the concentration is expressed in molality, it is expressed in moles of solute per kilogram of solvent. Since we are given the mass of the solvent, which is water, we can compute for the moles of solute NaNO3.

0.5 m = x mol NaNO3/0.5 kg water
x = 0.25 mol NaNO3

Since the molar mass of NaNO3 is 85 g/mol, the mass is

0.25 mol * 85 g/mol = 21.25 grams NaNO3 needed

4 0

3 years ago

Answer this if u know science please

scoundrel [369]

Answer:

1. DNA copying mistakes made during cell division, exposure to ionizing radiation, exposure to chemicals called mutagens, or infection by viruses.

2. both

Explanation: Since errors can occur when DNA replicates itself, mutations can occur in both asexually and sexually reproducing organisms.

5 0

3 years ago

How much positive charge is in 1.4 kg of oxygen? The atomic weight (15.9994 g) of oxygen contains Avogadro’s number of atoms, wi

Cerrena [4.2K]

Answer:

$6.7511\times 10^7\ C$

Explanation:

The atomic weight of oxygen = 15.9994 g

This mass corresponds to 1 mole of the oxygen atoms.

Thus,

15.9994 g mass of oxygen contains $6.02\times 10^{23}$ atoms of oxygen.

1.4 kg = 1400 g ( 1 kg = 1000 g)

So,

1400 g mass of oxygen contains $\frac {6.02\times 10^{23}}{15.9994}\times 1400$ atoms of oxygen.

Number of atoms in 1400 g of oxygen = $526.769754\times 10^{23}$

Also, 1 atom of oxygen contains 8 protons

Charge of 1 proton = + $1.602\times 10^{-19}\ C$

So, Charge on 1 atom of oxygen = $8\times 1.602\times 10^{-19}\ C$

Thus,

Charge on $526.769754\times 10^{23}$ atoms of oxygen = $526.94476\times 10^{23}\times 8\times 1.602\times 10^{-19}\ C=6.7533\times 10^7\ C$

Thus, positive charge in 1.4 kg of oxygen = $6.7511\times 10^7\ C$

5 0

3 years ago

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

jek_recluse [69]

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of $-6.4^oC$ and a molal freezing point depression constant $K_f=3.96^oC.kg/mol$ . A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at $-13.6^oC$ . Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

Answer: The mass of glycine that can be dissolved is $1.3\times 10^2g$

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

$\text{Freezing point of pure solvent}=\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $-6.4^oC$

Freezing point of solution = $-13.6^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_f$ = freezing point depression constant = $3.96^oC/m$

$m_{solute}$ = Given mass of solute (glycine) = ?

$M_{solute}$ = Molar mass of solute (glycine) = 75.07 g/mol

$w_{solvent}$ = Mass of solvent = 950. g

Putting values in equation 1, we get:

$-6.4-(-13.6)=1\times 3.96\times \frac{m_{solute}\times 1000}{75.07\times 950}\\\\m_{solute}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\m_{solute}=129.66g=1.3\times 10^2g$

Hence, the mass of glycine that can be dissolved is $1.3\times 10^2g$

5 0

3 years ago