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lana [24]
3 years ago
15

How much heat is needed to raise the temperature of a 5.0 gram block of aluminum from 22.0°c to 37.0°c?

Chemistry
2 answers:
Reptile [31]3 years ago
3 0
You'll need the specific heat capacity of aluminium to solve this question.
H=(0.005)(37-22)(specific heat capacity of aluminium)

dimulka [17.4K]3 years ago
3 0

Answer: 67.65 Joules

Explanation:

Q= m\times c\times \Delta T

Q= heat gained

m= mass of the substance = 5.0 g

c = heat capacity of aluminium = 0.902 J/g ° C      

\Delta T={\text{Change in temperature}}=(37-22)^oC=15^0C  

Q=5.0g\times 0.902J/g^oC\times 15^oC

Q= 67.65 Joules

Thus heat gained by 5.0 g of aluminum is = 67.65 Joules

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Consider the group a1 element sodium atomic number 11, the group 3a element aluminum atomic number 13, and the group 7a element
Akimi4 [234]
According to there chemical properties
7 0
3 years ago
Substances that are readily combustible or may cause fire through friction
Dvinal [7]

Answer :

Flammable substances

Explanation :

<em>Flammable substances</em> will catch fire and continue to burn when they contact an ignition source like a spark or a flame.

For example, <em>methanol</em> is a flammable liquid.

A flammable solid may also catch fire through friction. <em>Matches</em> are flammable solids.

3 0
3 years ago
What is the change in density if a sample goes from 3.21 g/L to 5.43 g/mL?
Step2247 [10]

Answer:

\Delta \rho =2.22 g/mL

Explanation:

Hello,

In this case, since a change in science is widely known to be considered as a subtraction between the the final and initial values of two measured variables and is represented via Δ, here the final density is 5.43 g/mL and the initial one was 3.21 g/mL, therefore, the change in density is:

\Delta \rho=\rho _f-\rho _i\\ \\\Delta \rho=5.43g/mL-3.21g/mL\\\\\Delta \rho =2.22 g/mL

Best regards.

3 0
3 years ago
Three different atoms or atomic cations with 3 electrons.
sveticcg [70]

Answer:

Li, Be^+, B^{2+}

Explanation:

To answer this question successfully, we need to remember that atoms are neutral species, since the number of protons, the positively charged particles, is equal to the number of electrons, the negatively charged particles. That said, we may firstly find an atom which has 3 electrons (and, as a result, 3 protons, as it should be neutral).

The number of protons is equal to the atomic number of an element. We firstly may have an atom with 3 protons and 3 electrons (atomic number of 3, this is Li).

Similarly, we may take the atomic number of 4, beryllium, and remove 1 electron from it. Upon removing an electron, it would become beryllium cation, Be^+.

We may use the same logic going forward and taking the atomic number of 5. This is boron. In this case, we need to remove 2 electrons to have a total of 3 electrons. Removal of 2 electrons would yield a +2-charged cation: B^{2+}.

6 0
3 years ago
Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
Montano1993 [528]

The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

<u>Answer:</u> The limiting reactant is ammonia (NH_3)

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For ammonia:</u>

Given mass of ammonia = 135.9 kg = 135900 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol

  • <u>For carbon dioxide gas:</u>

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol

The given chemical reaction follows:

2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = \frac{1}{2}\times 7994.12=3997.06mol of carbon dioxide

As, given amount of carbon dioxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ammonia is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reactant is ammonia (NH_3)

5 0
3 years ago
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