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ArbitrLikvidat [17]

3 years ago

9

Which power source directs the electrons from oxidation-reduction reactions to flow through a device to give the device power? O

A. An electric motor B. A battery C. An electromagnet O D. An electric generator

1 answer:

Lady bird [3.3K]3 years ago

4 0

Answer:

B. A battery.

Explanation:

I just took the test

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The brand names used on del monte, pillsbury, harley-davidson, and purina products are called _____ brands because of who owns t

patriot [66]

Manufacturers Is Your Answer

4 0

3 years ago

What is 67,890,000 in scientific notation?

Ulleksa [173]

I believe it's 6789 times 10^4. I could be wrong, due to the fact that I don't remember what's the definition of the scientific notion. I hope this helps, or at least gives you an idea! :D

7 0

3 years ago

Determine the molar mass of a compound that has a density of 0.1633 g/L at STP.<br> (show work)

hodyreva [135]

Answer:

M.Mass = 3.66 g/mol

Data Given:

M.Mass = M = ??

Density = d = 0.1633 g/L

Temperature = T = 273.15 K (Standard)

Pressure = P = 1 atm (standard)

Solution:

Let us suppose that the gas is an ideal gas. Therefore, we will apply Ideal Gas equation i.e.

P V = n R T ---- (1)

Also, we know that;

Moles = n = mass / M.Mass

Or, n = m / M

Substituting n in Eq. 1.

P V = m/M R T --- (2)

Rearranging Eq.2 i.e.

P M = m/V R T --- (3)

As,

Mass / Volume = m/V = Density = d

So, Eq. 3 can be written as,

P M = d R T

Solving for M.Mass i.e.

M = d R T / P

Putting values,

M = 0.1633 g/L × 0.08205 L.atm.K⁻¹.mol⁻¹ × 273.15 K / 1 atm

M = 3.66 g/mol

6 0

3 years ago

Thorium-234 has a half-life of 24.1 days. How many grams of a 150 g sample would you have after 120.5 days?

chubhunter [2.5K]

Answer:

8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days

Explanation:

The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.

We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.

Initial amount of Thorium-234 available as per the question is 150 grams

So now we start with 150 grams of Thorium-234

$150 \times \frac{1}{2}=24.1$

$75 \times \frac{1}{2} =48.2$

$34.5 \times \frac{1}{2} =72.3$

$17.25 \times \frac{1}{2} =96.4$

$8.625\times \frac{1}{2} =120.5$

So after 120.5 days the amount of sample that remains is 8.625g

In simpler way , we can use the below formula to find the sample left

$A=A_{0} \cdot \frac{1}{2^{n}}$

Where

$A_0$ is the initial sample amount

n = the number of half-lives that pass in a given period of time.

7 0

3 years ago

Synthesis<br> Decompisition<br> O Combustion<br> O Single Displacement<br> O Double Displacement

BabaBlast [244]

Answer:

Synthesis

Explanation:

Nothing else works

4 0

3 years ago