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3 years ago

10

2. Bart finds a solid shiny

1 answer:

notsponge [240]3 years ago

7 0

Answer:

1. Density of metal object = <em><u>11000 kg / m3</u></em>.

2. Could the object be gold = <em><u>No,because the density of gold is 19300 kg / m3</u></em>.

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A metal sample weighing 129.00 grams and at a temperature of 97.8 degrees Celsius was placed in 45.00 grams of water in a calori

Nitella [24]

Answer : The specific heat of metal is $0.481J/g^oC$ .

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of metal = 129.00 g

$m_2$ = mass of water = 45.00 g

$T_f$ = final temperature = $39.6^oC$

$T_1$ = initial temperature of metal = $97.8^oC$

$T_2$ = initial temperature of water = $20.4^oC$

Now put all the given values in the above formula, we get

$129.00g\times c_1\times (39.6-97.8)^oC=-45.00g\times 4.184J/g^oC\times (39.6-20.4)^oC$

$c_1=0.481J/g^oC$

Therefore, the specific heat of metal is $0.481J/g^oC$ .

4 0

4 years ago

Which of the following equations represents Photosynthesis? 16CO2 + 6H2O -----> C6H12O6 + 6O22C6H12O6 + 6O2 -----> CO2 + 6

lidiya [134]

Answer:

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

Explanation:

<em>Photosynthesis</em> is the chemical process carried out by plants for the conversion of inorganic matter (carbon dioxide and water) into organic matter (glucose) with the release of oxygen, using light (sun energy).

So the chemical process may be represented by:

<u>Word equation: </u>

carbon dioxide + water + sun energy → glucose + oxygen

<u>Skeleton equation:</u>

CO₂ + H₂O + sun energy → C₆H₁₂O₆ + O₂

<u>Balanced chemical equation:</u>

6CO₂ + 6H₂O + sun energy → C₆H₁₂O₆ + 6O₂

<u>Supressing the energy to show only the chemical compounds:</u>

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

8 0

3 years ago

Ham radio operators often broadcast on the 6-meter band. The frequency of this electromagnetic radiation is MHz. [speed of light

Volgvan

Answer:

50 MHz

Explanation:

The relation between frequency and wavelength is shown below as:

$c=frequency\times Wavelength$

c is the speed of light having value $3\times 10^8\ m/s$

Given, Wavelength = 6 m

Thus, Frequency is:

$Frequency=\frac{c}{Wavelength}$

$Frequency=\frac{3\times 10^8}{6}\ Hz$

$Frequency=50\times 10^{6}\ Hz$

Also,

Also, 1 Hz = $10^{-6}$ MHz

So,<u> Frequency = 50 MHz.</u>

8 0

3 years ago

The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?

anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

$K_a\times K_b=K_w$

$K_b$ for sodium formate is $K_b=\frac {K_w}{K_a}$

Given that:

$K_a$ of formic acid = $1.8\times 10^{-4}$

And, $K_w=10^{-14}$

So,

$K_b=\frac {10^{-14}}{1.8\times 10^{-4}}$

$K_b=5.5556\times 10^{-11}$

Concentration = 0.35 M

HCOONa ⇒ Na⁺ + HCOO⁻

Consider the ICE take for the formate ion as:

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻

At t=0 0.35 - -

At t =equilibrium (0.35-x) x x

The expression for dissociation constant of sodium formate is:

$K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}$

$5.5556\times 10^{-11}=\frac {x^2}{0.35-x}$

Solving for x, we get:

x = 0.44×10⁻⁵ M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

5 0

3 years ago

What two types of particles exist within an atomic nucleus?

MissTica

Protons and Neutrons, the electrons circle around the nucleus.
Hope this Helps!

6 0

4 years ago