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3 years ago

What is the identity of a 100. g sample of metal that, upon absorbing 4680 J of heat, increases in temperature by 52.0°C?

Chemistry

1 answer:

krok68 [10]3 years ago

3 0

The metal is aluminium

<u>Explanation:</u>

Given:

Heat, q = 4680 J

Mass, m = 100g = 0.1kg

ΔT = 52°C

sample = ?

We know:

q = mcΔT

On substituting the value we get:

$4680 = 0.1 X c X(52)\\\\c = \frac{4680}{5.2} \\\\c = 900 J/Kg^oC$

Thus, the metal is aluminium which has a specific heat capacity of 900 J/kg°C

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Calculate the molar mass of cacl2

kirza4 [7]

Answer:

$\boxed {\boxed {\sf 110.98 \ g/mol}}$

Explanation:

The molar mass is the mass of a substance in grams per mole.

To find it, add the mass of each element in the compound. These masses can be found on the Periodic Table.

The compound given is:

$CaCl_2$

The compound has 1 Ca (calcium) and 2 Cl (chlorine).

Mass of Calcium

The molar mass of calcium is 40.08 g/mol
There is only one atom of Calcium in CaCl₂, so the number above is what we will use.

Mass of Chlorine

The molar mass of chlorine is 35.45 g/mol
There are two atoms of chlorine in CaCl₂, therefore we need to multiply the molar mass by 2.
35.45 * 2= 70.9 g/mol

Molar Mass of CaCl₂

Now, to find the molar mass, add the molar mass of 1 calcium and 2 chlorine.
40.08 g/mol + 70.9 g/mol =110.98 g/mol

The molar mass of CaCl₂ is <u>110.98 grams per mole. </u>

6 0

3 years ago

Consider this reaction: 6 CO2 + 6 H2O + light equation C6H12O6 + 6 O2 If there were 2.38 x 102 g of H2O, 18.6 moles of CO2, and

alisha [4.7K]

H₂O would be the limiting reactant.

Balanced chemical equation:

6CO₂ + 6H₂O + light equation → C₆H₁₂O₆ + 6O₂

The amount of product that can be created is constrained by the reactant that is consumed first in a chemical reaction, commonly referred to as the limiting reactant (or limiting reagent).

Given

No. of moles of CO₂ = 18.6

Mass of H₂O = 2.38 × 10² g = 238g

No. of moles of H₂O = Given mass/ Molar mass

= 238 / 18 = 13.22 moles

Moles of H₂O = 13.22

According to the balanced chemical equation

6 moles of CO₂ react with 6 moles of H₂O

So the reactant that has less number of moles will be consumed first.

As the No. of moles of H₂O < No. of moles of CO₂

So, H₂O is the limiting reactant with 13.22 moles.

Hence, H₂O would be the limiting reactant.

Learn more about limiting reactant here brainly.com/question/14222359

#SPJ1

7 0

2 years ago

Which part of the atom is directly involved in chemical changes? A. the electron B. the energy levels C. the nucleus D. the char

7nadin3 [17]

The part of the atom that is involved in chemical changes is A. electron. The electrons that are in the most outer shells are called valence electrons which are easily removed or shared to form bonds. Valence electrons are related to the number of valence electrons

8 0

3 years ago

Can a bad thermostat cause radiator to blow

erastova [34]

I don't exactly know but probably so.

3 0

3 years ago

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.

sp2606 [1]

Answer:

Mass of CaCl₂ = 20 g

CaCO is presewnt in excess.

Mass of of CaCO₃ remain unreacted = 7.007 g

Explanation:

Given data:

Mass of calcium carbonate = 25 g

Mass of hydrochloric acid = 13.0 g

Mass of calcium chloride produced = ?

Chemical equation:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

Number of moles of CaCO₃= 25.0 g / 100.1 g/mol

Number of moles of CaCO₃ = 0.25 mol

Number of moles of HCl:

Number of moles of HCl = Mass /molar mass

Number of moles of HCl = 13.0 g / 36.5 g/mol

Number of moles of HCl = 0.36 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

CaCO₃ : CaCl₂

1 : 1

0.25 : 0.25

HCl : CaCl₂

2 : 1

0.36 : 1/2 × 0.36 = 0.18 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.18 mol × 110.98 g/mol

Mass of CaCl₂ = 20 g

The calcium carbonate is present in excess.

HCl : CaCO₃

2 : 1

0.36 : 1/2 × 0.36 = 0.18 mol

So, 0.18 moles react with 0.36 moles of HCl.

The moles of CaCO₃ remain unreacted = 0.25 -0.18

The moles of CaCO₃ remain unreacted = 0.07 mol

Mass of of CaCO₃ remain unreacted = Moles × molar mass

Mass of of CaCO₃ remain unreacted = 0.07 mol × 100.1 g/mol

Mass of of CaCO₃ remain unreacted = 7.007 g

7 0

3 years ago