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3 years ago

Help need help with this question answer plz 30 points

Chemistry

2 answers:

cestrela7 [59]3 years ago

8 0

Answer:

im not sure but i think its b

Explanation:

andrew-mc [135]3 years ago

8 0

I think it’s the second answer

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The process of_ causes rock to change compostion when reacting with oxgen

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Rusting causes rocks with metals in them to corrode and have changed compositions. The oxygen in the moisture in the air causes rock to rust.

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3 years ago

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Calculate the [oh−] in a solution with a ph of 12.52.

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PH + pOH = 14

12.52 + pOH = 14

pOH = 14 - 12.52

pOH = 1.48

[OH⁻] = 10^ -pOH

[OH⁻] = 10 ^- 1.48

[OH⁻] = 0.033 M

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3 years ago

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3 years ago

The burning of 48.7 g of zns in the presence of oxygen gives 220.0 kj of heat. what is h for the reaction as written below?

mr_godi [17]

Answer : The $\Delta H_{rxn}$ for the reaction is, 54.89 KJ

Solution : Given,

Mass of ZnS = 48.7 g

Molar mass of ZnS = 97.474 g/mole

$\Delta H=220KJ$

First we have to calculate the moles of ZnS.

$\text{ Moles of ZnS}=\frac{\text{ Mass of ZnS}}{\text{ Molar mass of ZnS}}=\frac{48.7g}{97.474g/mole}=0.499moles$

The balanced combustion reaction is,

$2ZnS+3O_2\rightarrow 2ZnO+2SO_2$

From the given reaction, we conclude that

As, 2 moles of ZnS gives energy = 220 KJ

So, 0.499 moles of ZnS gives energy = $\frac{220KJ}{2moles}\times 0.499moles=54.89KJ$

Therefore, the $\Delta H_{rxn}$ for the reaction is, 54.89 KJ

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3 years ago

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The percent composition by mass of an unknown compound with a molecular mass of 60.052 amu is 40.002% C, 6.7135% H, and 53.284%

Lemur [1.5K]

Answer: Empirical formula and molecular formula are $CH_2O$ and $C_2H_4O_2$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 40.002 g

Mass of H= 6.7135 g

Mass of O = 53.284 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{ 40.002g}{12g/mole}=3.3335moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{ 6.7135g}{1g/mole}=6.7135moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.284 g}{16g/mole}=3.3302moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.3335}{3.3302}=1$

For H = $\frac{6.7135}{3.3302}=2$

For O = $\frac{3.3302}{3.3302}=1$

The ratio of C : H : O= 1 : 2 : 1

Hence the empirical formula is $CH_2O$

The empirical weight of $CH_2O$ = 1(12)+2(1)+1(16)= 30 g.

The molecular weight = 60.052 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{ 60.052 }{30}=2$

The molecular formula will be= $2\times CH_2O=C_2H_4O_2$

Thus empirical formula and molecular formula are $CH_2O$ and $C_2H_4O_2$

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3 years ago