Where are alkaline earth metals found on the periodic table?

Why would you expect sodium (Na) to react strongly with chlorine (Cl)?

Snowcat [4.5K]

3. is the answer, <span>Sodium needs to lose one electron, and chlorine needs to gain one electron. This is because Sodium's row always wants to give away an electron, while Chlorine's row wants to gain an electron.</span>

5 0

3 years ago

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4. Calculate the total resistance for two 180 ohm resistors connected in<br> parallel

solmaris [256]

Answer:

90 ohms

Explanation:

1/r = 1/180 + 1/180

1/r= 2/180

take the reciprocal of 2/180 which is 180/2 and its 90 ohms

3 0

3 years ago

child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s

Strike441 [17]

Answer:

θ = 13.7º

Explanation:

According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

$\Delta K + \Delta U = W_{nc} (1)$

ΔK, is the change in kinetic energy, as follows:

$\Delta K = \frac{1}{2}* m* (v_{f} ^{2} - v_{0} ^{2}) (2)$

ΔU, is the change in the gravitational potential energy.
If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

$\Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)$

Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

$W_{nc} = F_{f} * d * cos 180\º \\\\ = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)$

Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

$\frac{1}{2}* (v_{f} ^{2} - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d$

Replacing by the givens and the knowns, we can solve for sin θ, as follows: $\frac{1}{2}*( (4.30 m/s) ^{2} - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2} = 0.236$ ⇒ θ = sin⁻¹ (0.236) = 13.7º

8 0

3 years ago

The specific heat of water in its solid phase (ice) is 2090 J/(kg K), while in the liquid phase (water) its specific heat is 419

oee [108]

Answer:

d. 149 ⁰C.

Explanation:

Given;

mass of the block of ice, m = 2 kg

specific heat capacity of the ice, C = 2090 J/(kgK)

initial temperature of the ice, t₁ = -90 ⁰C

heat added to the ice, H = 1,000,000 J

let the final temperature of the ice = t₂

The final temperature of the ice after adding the heat is calculated as follows;

$H = mC_{ice} \Delta t\\\\H = mC_{ice} (t_2 - t_1)\\\\1,000,000 = 2 \times 2090 \times (t_2 - (-90))\\\\1,000,000 = 4,180(t_2 +90)\\\\1,000,000 = 4,180t_2 + 376,200\\\\1,000,000 - 376,200 = 4,180t_2\\\\623,800 = 4,180 t_2\\\\t_2 = \frac{623,800}{4,180} \\\\t_2 = 149 \ ^0C$