Chandra forms her question: "What is the effect of soda on plant growth?" What would be a good hypothesis for this question?

A 0.300 g piece of copper is heated and fashioned into a bracelet. The amount of energy transferred by heat to the copper is 66,

Alja [10]

Answer:

X=600 degrees celsius

Explanation:

We can evaluate by using the equation ==> Q=m(c)(delta T)

-We know the mass is 0.3 g

-Q= 66,300 J

-C= 390 j/g 0C

-Therefore, we need to solve for delta T (which will be represented by x)

Q= mct

66,300 J= 0.3 g (390 J/g 0C) (t)

The unit of measurement (g-grams) can be crossed out on the left side of our expression.

66,300 J=0.3(390 J 0C) (t)

-If we convert into a proportional relationship:

66,300 J 117 J/0C x

------------ = -----------

117 J/0C 117 J/0C

-Joules on both sides can be crossed off

*117 J/0C came from the product in terms of 0.3(390 J 0C)

*If we find the quotient of 66,300 J and 117 J/0C, we get 566.6 repeating

-You may keep the answer as it is, although for simplicity reasons, we can round to the nearest hundred.

-Therefore, x=600.

-----Remember, x represented the delta t in our equation. (I made notice of that when evaluating).

Hope this helps! :)

3 0

1 year ago

. Calculate the mass of O2 produced if 3.450 g potassium chlorate is completely decomposed by heating in presence of a catalyst

Vesnalui [34]

2 KClO3(s) → 3 O2(g) + 2 KCl(s)

Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates.

molar mass of KClO3 = 122.5
Moles of KClO3 = 3.45 / 122.55 = 0.028

Moles of O2 produce = $\frac{3}{2} \times 0.028$

= 0.042 moles

molar mass of O2 = 32

so, mass of O2 = 32 x 0.042 = 1.35 g

5 0

2 years ago

Which of the following statements best describes the properties of aluminum-28?

Contact [7]

d. When aluminum-28 undergoes beta decay, silicon-28 is produced.

Explanation:

When the nuclei of aluminium-28 decays, it produces silicon- 28:

Aluminium ²⁸₁₃Al

Silicon 28 ²⁸₁₄Si

beta particle ⁰₋₁ $\beta$

²⁸₁₃Al → ²⁸₁₄Si + ⁰₋₁ $\beta$

This way, the mass and atomic number are conserved.

Conservation of mass number:

28 = 28 + 0, 28 = 28

13 = 14 -1 , 13 = 13

Learn more:

Balancing nuclear equations brainly.com/question/10094982

#learnwithBrainly

5 0

2 years ago

In the PhET simulation, make sure that the checkbox Stable/Unstable in the bottom right is checked. Using the PhET simulation as

lord [1]

Answer:

kindly check the EXPLANATION SECTION

Explanation:

In order to be able to answer this question one has to consider the neutron proton ratio. Considering this ratio will allow us to determine the stability of a nuclei. The most important rule that helps us in determination of stability is that when the Neutron- Proton ratio of any nuclei ranges from to 1 to 1.5, then we say the nuclei is STABLE.

Also, we need to understand that when the Neutron- Proton ratio is LESS THAN 1 or GREATER THYAN 1.5, then we say the nuclei is UNSTABLE.

So, let us check which is stable and which is unstable:

a. 4 protons and 5 neutrons = Neutron- proton ratio = N/P = 5/4= stable.

b. 7 protons and 7 neutrons = Neutron- proton ratio = N/P = 7/7= 1 = stable.

c. 2 protons and 3 neutrons = Neutron- proton ratio = N/P = 3/5 =0.6 =unstable.

d. 3 protons and 0 neutrons = Neutron- proton ratio = N/P = 0/3= 0= unstable.

e. 6 protons and 5 neutrons = Neutron- proton ratio = N/P = 5/6= 0.83 = unstable.

f. 9 protons and 9 neutrons = Neutron- proton ratio = N/P = 9/9 = 1 = stable.

g. 8 protons and 7 neutrons = Neutron- proton ratio = N/P = 7/8 =0.875 = unstable.

h. 1 proton and 0 neutrons = Neutron- proton ratio = N/P = 0/1 =0 = unstable

6 0

2 years ago

Given that the rate constant is 4.0×10−4 m−1 s−1 at 25.0 ∘c and that the rate constant is 2.6×10−3 m−1 s−1 at 42.4 ∘c, what is t

stepladder [879]

So here's how you find the answer:

Given: (rate constants)

K₁ = 4.0 x 10⁻⁴ M⁻¹s⁻¹.
T₁ = 25.0 C = 293 K.
k₂ = 2.6 x10⁻³ M⁻¹s⁻¹.
T₂ = 42.4 C = 315.4 K.
R = 8,314 J/K·mol.

Use the equation:

ln(k₁/k₂) = Ea/R (1/T₂ - 1/T₁).

Transpose:

Ea = R·T₁·T₂ / (T₁ - T₂) · ln(k₁/k₂)

Substitute within the given transposed equation:

Ea = 8,314 J/K·mol · 293 K · 315.4 K ÷ (293 K - 315.4 K) · ln (4.0 x 10⁻⁴ M⁻¹s⁻¹/ 2.6 x 10⁻³ M⁻¹s⁻¹).

Continuing the solution we get:

Ea = 768315 J·K/mol ÷ (-22,4 K) * (-1.87)

The value of EA is:

Ea = 64140.58 J/mol ÷ 1000 J/kJ = 64.140 kJ/mol.

4 0

2 years ago