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3 years ago

PLZ HELP THERE IS A FILE ATTACHED WILL GIVE U BRAINLIEST IF U HELP

Chemistry

2 answers:

erastovalidia [21]3 years ago

6 0

Answer:

I think C (Might not be true)

Explanation:

Svetllana [295]3 years ago

5 0

Answer:

Explanation:

Should be D as it follows the pattern on the period table in group 1.

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Explain why anhydrous aluminium chloride is fairly soluble in organic solvent while anhydrous magnesium chloride is insoluble

prohojiy [21]

Aluminium chloride is covalent hence soluble in organic solvent while magnesium chloride being ionic is insoluble in organic solvent

8 0

3 years ago

Save the ________ used to extract metals.

MrRa [10]

Answer:

Save the *minerals* used to extract metals.

6 0

2 years ago

The concentration of Rn−222 in the basement of a house is 1.45 × 10−6 mol/L. Assume the air remains static and calculate the con

bonufazy [111]

<u>Answer:</u> The concentration of radon after the given time is $3.83\times 10^{-30}mol/L$

<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

We are given:

$t_{1/2}=3.82days$

Putting values in above equation, we get:

$k=\frac{0.693}{3.82}=0.181days^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $0.181days^{-1}$

t = time taken for decay process = 3.00 days

$[A_o]$ = initial amount of the reactant = $1.45\times 10^{-6}mol/L$

[A] = amount left after decay process = ?

Putting values in above equation, we get:

$0.181days^{-1}=\frac{2.303}{3.00days}\log\frac{1.45\times 10^{-6}}{[A]}$

$[A]=3.83\times 10^{-30}mol/L$

Hence, the concentration of radon after the given time is $3.83\times 10^{-30}mol/L$

7 0

3 years ago

Information-

tekilochka [14]

Answer:

So first thing to do in these types of problems is write out your chemical reaction and balance it:

Mg + O2 --> MgO

Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.

To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.

The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.

The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.

Percent yield is acutal/theoretical, .66/.693, or 95.24%.

I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.

Hope this helps.

6 0

2 years ago

The equation for the formation of water from hydrogen gas and oxygen gas is 2H2+O2→2H2O.

inn [45]

Answer:

180g

Explanation:

H:1 O:16

2H2+O2 → 2H2O

2 2(16) 2(1)+(16)

32 18

Now,

32g of O → 2(18)g of H2O

160g of O → 2(18)g divides by 32g times 160g

=180g

5 0

2 years ago