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3 years ago

Help plz:)))I’ll mark u Brainliest

Chemistry

1 answer:

svet-max [94.6K]3 years ago

8 0

I’m not sure if this s what your asking but here’s the formula, 6co2 + 6h2o + suns energy + c6h12o6 + 6o2

Carbon dioxide + water + Sun’s energy + Sugar (Glucose) + Oxygen

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What particle has almost mo mass

coldgirl [10]

Photon has no mass at all

6 0

3 years ago

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Which of the following possess the greatest concentration of hydroxide ions?

jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

(a) a solution of pH 3.0

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$11=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-11}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-11}M$

(b) a solution of 0.10 M $NH_3$

As we know that 1 mole of $NH_3$ is a weak base. So, in a solution it will not dissociates completely.

So, the $OH^-$ concentration will be less than 0.10 M

(c) a solution with a pOH of 12.

We have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$12=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-12}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-12}M$

(d) a solution of 0.10 M NaOH

As we know that $NaOH$ is a strong base. So, it dissociates to give $Na^+$ ion and $OH^-$ ion.

So, 0.10 M of $NaOH$ in a solution dissociates to give 0.10 M of $Na^+$ ion and 0.10 M of $OH^-$ ion.

Thus, the $OH^-$ concentration is, 0.10 M

(e) a $1\times 10^{-4}M$ solution of $HNO_2$

As we know that 1 mole of $HNO_2$ in a solution dissociates to give 1 mole of $H^+$ ion and 1 mole of $NO_2^-$ ion.

So, $1\times 10^{-4}M$ of $HNO_2$ in a solution dissociates to give $1\times 10^{-4}M$ of $H^+$ ion and $1\times 10^{-4}M$ of $NO_2^-$ ion.

The concentration of $H^+$ ion is $1\times 10^{-4}M$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.0\times 10^{-4})$

$pH=4$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$10=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-10}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-10}M$

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

3 0

4 years ago

How many molecules of co2,h2o,c3h8, and o2 will be present if the reaction goes to completion?

Nadya [2.5K]

Let initially there are 10 molecules of O2 and 3 molecules of C3H8 present

The reaction will be

C3H8(g) + 5O2(g) ----> 3CO2(g) + 4H2O

so here oxygen molecules are limiting as for 3 molecules of C3H8 we need 15 molecules of O2

now the given 10 molecules of O2 will react with only 2 molecules of C3H8 and they will form six molecules of CO2 and 8 molecules of H2O

Hence answer is

molecules of CO2 formed = 6

Molecules of H2O formed = 8

molecules of C3H8 left = 1

molecules of O2 left = 0

7 0

3 years ago

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18. What is the percent by mass sugar of a solution made by dissolving 12.45 grams of sugar in 100 grams of water?

Fantom [35]

The percent by mass sugar of a solution : 11.07%

<h3>Further explanation</h3>

Given

mass of sugar = 12.45 g

mass of water = 100 g

Required

The percent by mass

Solution

Mass of solution :

$\tt mass~sugar+mass~water=12.45+100=112.45~g$

Percent mass of sugar :

$\tt \%mass=\dfrac{mass~sugar}{mass~solution}\times 100\%\\\\\%mass=\dfrac{12.45}{112.45}\times 100\%=11.07\%$

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3 years ago

Mg(ClO2)2 and Al(ClO2)3 contain the same number of atoms

Vladimir [108]

Mg(ClO2)2 has an atom of magnesium, 2 moles of chlorine and 4 moles of oxygen. Al(ClO2)3, on the other hand, is composed of one mole of aluminum, one mole of chlorine and six moles of oxygen. The number of molecules of both is equal to 6.022 x10^23 molecules per mole of the compound.

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3 years ago