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3 years ago

At STP, a 50-gram sample of H20(I) and a 100-gram sample of H20(I) have

Chemistry

2 answers:

Katyanochek1 [597]3 years ago

3 0

Answer:

(1) the same chemical properties .

Explanation:

Hello!

In this case, among the options:

(1) the same chemical properties

(2) the same volume

(3) different temperatures

(4) different empirical formulas

We can see that they have the same chemical properties as they at the same conditions, same type of bond (polar), molecular geometry, bond angle (104.5 °) and so on. Nevertheless, at STP (1 atm and 273.15 K) they do not have the same volume since the larger the mass, the larger the volume, they have the same temperature and the both of them are H₂O.

It means that the answer is (1) the same chemical properties .

Best regards.

Paladinen [302]3 years ago

3 0

Good answer man <3

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A dissolved mixture when it mixed together hope this helps

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3 years ago

Read 2 more answers

When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3), which causes acid rain

Gekata [30.6K]

Answer:

$n_{HNO_3}=0.584molHNO_3$

Explanation:

Hello.

In this case, given the described reaction, the formation of nitric acid turns out:

$NO_2+H_2O\rightarrow HNO_3$

Since two hydrogen atoms are present at the reactants we balance it as follows:

$3NO_2+H_2O\rightarrow 2HNO_3+NO$

In such a way, since there is a 1:2 mole ratio between water and nitric acid, the produced moles of nitric acid, turns out:

$n_{HNO_3}=0.292molH_2O*\frac{2molHNO_3}{1molH_2O}\\\\ n_{HNO_3}=0.584molHNO_3$

Best regards!

5 0

3 years ago

A compound decomposes by a first-order process. What is the half-life of the compound if 25.0% of the compound decomposes in 60.

amid [387]

Answer : The half-life of the compound is, 145 years.

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant = ?

t = time passed by the sample = 60.0 min

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process = 100 - 25 = 75 g

Now put all the given values in above equation, we get

$k=\frac{2.303}{60.0}\log\frac{100g}{75g}$

$k=4.79\times 10^{-3}\text{ years}^{-1}$

Now we have to calculate the half-life of the compound.

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

$4.79\times 10^{-3}\text{ years}^{-1}=\frac{0.693}{t_{1/2}}$

$t_{1/2}=144.676\text{ years}\approx 145\text{ years}$

Therefore, the half-life of the compound is, 145 years.

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4 years ago

What are the rules for writing

Alborosie

Answer:

When naming molecular compounds prefixes are used to dictate the number of a given element present in the compound. ” Mono-” indicates one, “Di-” indicates two, “Tri-” is three, “Tetra-” is four, “Penta-” is five, and “Hexa-” is six, “Hepta-” is seven, “Octo-” is eight, “Nona-” is nine, and “Deca-” is ten.

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3 years ago

Suppose that 2.14 grams of H2SO4 is mixed with enough water to make 225 mL of solution. Determine the Molarity (M) of the soluti

Alik [6]

Answer:

0.0970 M

Explanation:

Remember this equation:

mol/M x V

Convert it so that you can get M.

M=mol/V

Convert the 2.14 grams of H2SO4 into mols

=0.0218

Convert mL to L

225/1000

=0.225

Plug it in.

0.0218/0.225

=0.0970 M

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3 years ago