Answer:
The mass of 10 cm³of a 0.4 g/dm³ solution of sodium carbonate is 0.004 grams
Explanation:
The question is with regards to density calculations
The density of the given sodium carbonate solution, ρ = 0.4 g/dm³
The volume of the given solution of sodium carbonate, V = 10 cm³ = 0.01 dm³
Therefore, we have;
The mass, "m", of the sodium carbonate in = ρ×V = 0.4 g/dm³ × 0.01 dm³ = 0.004 g
The mass of 10 cm³ (10 cm³ = 0.01 dm³) of a 0.4 g/dm³ solution of sodium carbonate, m = 0.004 g.
Here is 5
Dissolved Load - elements dissolved in solution
Suspended Load - very fine grained sediment such as clay and silt carried in suspension. The size grains that can be carried in suspension are dependent on the current velocity
Wash Load - a subset of the suspension load, extremely small particles (clay) that will remain in suspension independent of turbulence in the river
Saltation Load - particles that are temporarily carried in suspension but move by bouncing along the bottom
<span>Bed Load - sediment that moves by rolling or sliding along the bottom. These are generally the coarser grained sediments such as sand and gravel.</span>
Hi there! Air and sunlight can definitely be reused. Those are abundant and renewable resources. Therefore, A and D are eliminated. There is a limited amount of water, however, it's impossible to run out of it to the point that there's no more on Earth. C is out. The only answer choice that makes sense is coal, because it's a nonrenewable resource, and it takes millions of years to make more of. It's a fossil fuel, so once we use them up, we can't get anymore during our lives. The answer is B: coal.
The answer is _Dissolving_.
Hope tis helps
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
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
Where
is the initial sample amount
n = the number of half-lives that pass in a given period of time.
