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2 years ago

When 5800 joules of energy are applied to a 15.2-kg piece of lead metal, how much does the temperature change by

2 answers:

8 0

<span>Heat gained or absorbed in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. The heat capacity of aluminum at 25 degrees celsius is 0.9 J/g-C. It is expressed as follows:</span><span>

Heat = mC(T2-T1)
5800 J = 152000(0.90)(</span>ΔT)

ΔT = 0.42 °C change in temperature

alexgriva [62]2 years ago

5 0

Answer: 2.98 K

Explanation:

$Q= m\times c\times \Delta T$

Q= heat gained or released= 5800 J

m= mass of the substance = 15.2 kg = 15200 g (1kg=1000g)

c = heat capacity of lead= 0.128 J/gK

$\Delta T={\text{Change in temperature}}=?$

$5800J=15200g\times 0.128J/gK\times \Delta T$

$\Delta T= 2.98K$

Thus change in temperature is 2.98 K.

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Which of the following is the best description of the results of a chemical reaction?

dedylja [7]

Answer:

A, new substance is formed with different properties

Explanation:

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3 years ago

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than in $Ag_3PO_4$ $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

To determine the concentration of $[Ca^+]$ when the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

Therefore;

the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440 × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$ remaining = 12.86%

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3 years ago

How do her cells use these molecules to release energy for her body to run?

Anika [276]

In cells use oxygen to release energy stored in sugars such as glucose. In fact, most of the energy used by the cells in your body is provided by cellular respiration. Just as photosynthesis occurs in organelles called chloroplasts, cellular respiration takes place in organelles called mitochondria.

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2 years ago

3. A 0.025L solution of HCl is neutralized by 0.018L of a 1.0 M NaOH solution. What is the concentration of the HCl solution?

Nookie1986 [14]

The concentration of the HCl solution is 0.72 M.

<h3>How do we calculate the concentration?</h3>

Concentration of the required solution by the use of the known concentration solution will be determine by using the below equation as:

M₁V₁ = M₂V₂, where

M₁ & V₁ are the molarity and volume of the HCl solution.
M₂ & V₂ are the molarity and volume of the NaOH solution.

On putting values in the above equation, we get

M₁ = (1)(0.018) / (0.025) = 0.72 M

Hence required concentration of HCl is 0.72M.

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1 year ago

The stream table shows the time needed for water to soak into the playfield soil.

Sloan [31]

Answer:

TRUE

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2 years ago