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3 years ago

Which is an example of a chemical reaction? A) melting ice B) rusting iron C) water evaporating D) rubbing a marker on paper

Chemistry

2 answers:

zvonat [6]3 years ago

8 0

Necklaces of seashells, lion teeth, and beer claws. Also sculpture of animals.

Ksju [112]3 years ago

4 0

B rusting iron because whatever is making it rust is reacting with the iron to cause it to rust.

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Pani-rosa [81]

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3 years ago

Read 2 more answers

Calculate by (a)% weight and (b) %mole each of the elements present in sugar

Musya8 [376]

Explanation:

Molecular mass of sugar = $C_{12}H_{22}O_{11}$ : = 432 g/mol

Atomic mass of carbon atom = 12 g/mol

Atomic mass of hydrogen atom = 1 g/mol

Atomic mass of oxygen atom = 16 g/mol

a) Percentage of an element in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of compound}}\times 100$

Percentage of carbon by weight in $C_{12}H_{22}O_{11}$ :

$\frac{12\times 12 g/mol}{342 g/mol}\times 100=42.10\%$

Percentage of hydrogen by weight in $C_{12}H_{22}O_{11}$ :

$\frac{22\times 1g/mol}{342 g/mol}\times 100=6.43\%$

Percentage of oxygen by weight in $C_{12}H_{22}O_{11}$ :

$\frac{11\times 16g/mol}{342 g/mol}\times 100=51.46\%$

b) Percentage of mole each of the elements present in sugar:

= $\frac{\text{Moles of atoms of an element}}{\text{total moles of all types of atoms}}\times 100$

In mole of sugar we have 12 moles of carbon atom , 22 moles of hydrogen atoms and 11 moles of oxygen atoms.

Percentage of carbon by mole in $C_{12}H_{22}O_{11}$ :

$\frac{12 mol}{45 mol}\times 100=26.66\%$

Percentage of hydrogen by mole in $C_{12}H_{22}O_{11}$ :

$\frac{22 mol}{45 mol}\times 100=48.88\%$

Percentage of oxygen by mole in $C_{12}H_{22}O_{11}$ :

$\frac{11 mol}{45 mol}\times 100=24.44\%$

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4 years ago

In which situation is it better to have low friction?

Ray Of Light [21]

A student walks down a hall I’m pretty sure

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3 years ago

A balloon has 18.9 moles of hydrogen gas at a volume of 428 L. What would the volume of the gas be if the balloon had 14.0 moles

den301095 [7]

Answer:

V₂ = 317 L

Explanation:

Given data:

Initial number of moles of hydrogen = 18.9 mol

Initial volume of gas = 428 L

Final volume = ?

Final number of moles = 14.0 mol

Solution:

According to the Avogadro law,

Number of moles of gas is directly proportional to the volume of gas at constant temperature and pressure.

Mathematical expression:

V₁/n₁ = V₂/n₂

V₁ = Initial Volume of balloon

n₁ = initial number of moles

V₂ = Final volume of balloon

n₂ = Final number of moles

Now we will put the values.

428 L / 18.9 mol = V₂/ 14 mol

V₂ =428 L × 14 mol / 18.9 mol

V₂ = 5992 L /18.9

V₂ = 317 L

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4 years ago

Combustion analysis of 63.8 mg of a c, h and o containing compound produced 145.0 mg of co2 and 59.38 mg of h2o. what is the emp

Makovka662 [10]

<span>: The empirical formula for the compound is C3H60 (see below) CO2 is the only product containing C, C produced = 145.0 mg CO2 x (1 g / 1000 mg) x (1 mole CO2 / 44.0 g CO2) x (1 mole C / 1 mole CO2) = 0.00330 moles C. H2O is the only product containing H, H produced = 59.38 mg H2O x (1 g / 1000 mg) x (1 mole H2O / 18.0 g H2O) x (2 moles H / 1 mole H2O) = 0.00660 moles H. Oxygen is in both and the unknown reacts with oxygen(in the air) 0.00330 moles C x (12.0 g C / 1 mole C) = 0.0396 g C = 39.6 mg C 0.00660 moles H x (1.01 g H / 1 mole H) = 0.00667 g H = 6.7 mg H Because the unknown weighed 63.8 mg and consists off justC, H, and O, then mass O = g unknown - g C - g H = 63.8 mg - 39.6 mg - 6.7 mg = 17.5 mg = 0.0175 g 0.0175 g O x (1 mole O / 16.0 g O) = 0.00109 moles O The mole ratio of C:H:O is: C = 0.00330 H = 0.00660 O = 0.00109 Divide by the smallest you get: C = 0.00330 / 0.00109 = 3.03 H = 0.00660 / 0.00109 = 6.06 O = 0.00109 / 0.00109 = 1.00</span>

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4 years ago