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Grace [21]
3 years ago
14

The layered structures made of calcium carbonate by precambrian cyanobacteria are called _____.

Chemistry
2 answers:
Valentin [98]3 years ago
7 0
<span>stromatolites I think
</span>
Misha Larkins [42]3 years ago
6 0

Im pretty sure its Trilobites man.

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What is the formula for dinitrogen pentoxide?<br> A)NO5<br> B)N5O2<br> C)N2O5<br> D)N2O7
Juliette [100K]

Answer:

nick gur

Explanation:

5 0
2 years ago
Read 2 more answers
Determine the volume of atmospheric air (at 14 lb/in.^2) needed to fill your bike tires (assuming it holds 500 mL of air) to the
Fiesta28 [93]

Boyle Law says “the pressure of fixed amount of ideal gas which is at constant temperature is inversely proportional to its volume".<span>

P = 1/V

<span>Where, P is pressure of the ideal gas and V is volume of the ideal gas.</span>

<span>For two situations, this law can be added as;
P</span>₁V₁ = P₂V₂<span>

</span><span>14 lb/in² x V₁ = 70 lb/in² x 500 mL</span><span>
                 </span><span>V₁ = 2500 mL</span><span>

Hence, the needed volume of atmospheric air = 2500 mL

<span>Here, we made two </span>assumptions. They are,
1. The atmospheric air acts as ideal gas.
2. Temperature is a constant.

<span>We didn't convert the units to SI units since converting volume and pressure are products of two numbers, they will cut off. </span></span></span>

3 0
3 years ago
A fixed mass of oxygen gas occupies 300cm cube at 0 degree centigrade. what volume would the gas occupy at 15 degree centigrade​
egoroff_w [7]

Answer:

Volume occupied by oxygen gas at 15 degree centigrade​ is equal to 316.5 centimeter cube

Explanation:

Assuming Pressure is constant.

\frac{V_1}{T_1} = \frac{V_2}{T_2}

where T1 and T2 are temperature in Kelvin

Substituting the give values we get-

\frac{300}{273} = \frac{V_2}{288}

V_2 = \frac{288*300}{273} \\V_2 = 316.5

Volume occupied by oxygen gas at 15 degree centigrade​ is equal to 316.5 centimeter cube

5 0
2 years ago
Pls help one chemistry question :/
jolli1 [7]

Answer:

the anserw should be 665KJ

8 0
2 years ago
A 98.0°C piece of cadmium (c=.850J/g°C) is placed in 150.0g of 37.0°C water. After sitting for a few minutes, both have a temper
Eduardwww [97]
The answer is 19.9 grams cadmium. 
Assuming there was no heat leaked from the system, the heat q lost by cadmium would be equal to the heat gained by the water:
     heat lost by cadmium = heat gained by the water
     -qcadmium = qwater
Since q is equal to mcΔT, we can now calculate for the mass m of the cadmium sample:
     -qcadmium = qwater
     -(mcadmium)(0.850J/g°C)(38.6°C-98.0°C)) = 150.0g(4.18J/g°C)(38.6°C-37.0°C)
     mcadmium = 19.9 grams
3 0
2 years ago
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