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REY [17]
2 years ago
13

Measuring Volume

Chemistry
1 answer:
Dvinal [7]2 years ago
7 0
Seems right, but if it’s just asking for one i would pick B
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An iron ore was found to contain 92.0% pyrite (fool's gold). Pyrite has the formula FeS2and is 46.5% iron. How many grams of iro
Liula [17]
Firstly calculate the grams in the last 8 percent before moving onto the pyrite section.
50.8x0.08=4.064g
We know that iron ore in this case has 92 percent pyrite which contains 46.5 percent iron so we do 50.8x0.92=46.736g from this we need to find 46.5 percent of the iron content in the 92 percent pyrite section then add this answer to the 8 percent of iron ore we found at the start.46.736x0.465=21.73224g
21.73224g+4.064=25.79624g of iron ore 25.8g(3sf)
4 0
3 years ago
Write a balenced chemical equation for the reaction: include abbreviation for the physical states-
WITCHER [35]

Answer:

MgBr(aq) + (NH4)3PO4(aq) -------> NH4Br(aq) + Mg3(PO4)2(s)

Explanation:

4 0
3 years ago
Assuming it behaves as an ideal gas, calculate the density of sulfur dioxide, so2, at stp.
Arisa [49]
Hello!

At Standard Pressure and Temperature, an ideal gas has a molar density of  0,04464 mol/L.

So, we need to apply a simple conversion factor to calculate the density of Sulfur Dioxide using the molar mass of Sulfur Dioxide.

\frac{0,04464 mol SO_2}{1 L SO_2}* \frac{64,066 g SO_2}{1 mol SO_2}=2,8599 g/L

So, the Density of Sulfur Dioxide (SO₂) at STP is 2,8599 g/L

Have a nice day!


6 0
3 years ago
Read 2 more answers
Greater than ___________ mcg/dl is considered a high level of morning cortisol.
Natalija [7]

Greater than 23mcg/dl is considered a high level of morning cortisol.

Normally, cortisol levels rise during the early morning hours and are highest about 7AM. They drop very low in the evening and during the early phase of sleep. If you do not have this daily change (diurnal rhythm) in cortisol levels, you may have overactive adrenal glands. This condition is called Cushing's syndrome.

5 0
3 years ago
What is the change in internal energy ( ΔU ) of the system if q = –8 kJ and w = –1 kJ for a certain process?
Radda [10]

Answer:

Change in internal energy (ΔU) = -9 KJ

Explanation:

Given:

q = –8 kJ [Heat removed]

w = –1 kJ [Work done]

Find:

Change in internal energy (ΔU)

Computation:

Change in internal energy (ΔU) = q + w

Change in internal energy (ΔU) = -8 KJ + (-1 KJ)

Change in internal energy (ΔU) = -8 KJ - 1 KJ

Change in internal energy (ΔU) = -9 KJ

6 0
3 years ago
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