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Nikolay [14]
3 years ago
10

One mole of which of these compounds contains two moles of hydrogen atoms?NaOH H2S NH3 CH4

Chemistry
1 answer:
Nookie1986 [14]3 years ago
8 0

Answer:

H2S.

Explanation:

That would be H2S,  hydrogen sulphide. (Smells like bad eggs!).

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Determining Density and Using Density to Determine Volume or Mass
Shalnov [3]

corrected question:

Determining Density and Using Density to Determine Volume or Mass

(a) Calculate the density of mercury if 1.00 × 10 g occupies a volume of 7.36 cm³

(b) Calculate the volume of 65.0 g of liquid methanol (wood alcohol) if its density is 0.791 g/mL.

(c) What is the mass in grams of a cube of gold (density = 19.32 g/cm) if the length of the cube is 2.00 cm?

(d) Calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm³ A student needs 15.0 g of ethanol for an experiment. If the density of ethanol is 0.789 g/mL, how many milliliters of ethanol are needed? What is the mass, in grams, of 25.0 mL of mercury (density = 13.6 g/mL)?

Answer:

density = \frac{mass}{volume}

ρ=m/v ,m=ρv,    v=m/ρ

(a)m=1*10g  , v=7.36cm³

    ρ=10/7.36 =1.36g/cm³

(b) m=65g, ρ=0.791 g/mL.

   v= 65/0.791 =82.17g/mL

(c) ρ=19.32g/cm³, l=2cm, v=l³=8cm³

    m=19..32*8=154.56g/cm³

(d) mass of copper=374.5g , v=41.8cm³

   ρ=374.5/41.8 =8.96g/cm³

 mass of ethanol=15g,  density of ethanol=0.789g/mL

v=15/0.789 =19.01mL

volume of mecury=25mL, density of mercury=13.6g/mL

m=25*13.6=340g

4 0
3 years ago
A solution is prepared by mixing 38.0L of pyrimidine and 442.0L of water. What is the percent volume of pyrimidine in this solut
klasskru [66]

I have the same question and cant still answer it so I need the answers

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Which term best describes the role of hydrogen gas in the formation of water molecules?
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An atom can be compared to our solar system because: WILL MAKE BRAINLIEST!
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Answer:

uh i think its D All of the above

Explanation:

sorry if its wrong

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3 years ago
The combustion reaction of c3h8o with o2 is represented by this balanced chemical equation. 2c3h8o + 9o2 6co2 + 8h2o when 3.00 g
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<span>According to the question- 1 mol C3H8O = 60.096 g C3H8O 2 mol C3H8O = 9 mol O2 1 mol O2 = 31.998 g O2 [(3.00 g C3H8O)/1][(1 mol C3H8O)/(60.096)][(9 mol O2)/(2 mol C3H8O)][(32.998 g O2)/(1 mol O2)] = 7.1880435 g O2 Since 7.1880435 g of O2 is needed, and 7.38 g of O2 is available, 0.199565 g of O2 will be left over and oxygen is present in excess. Next, we need to convert 0.199565 g of O2 into moles of O2: [(0.199565 g O2)/1][(1 mol O2)/(31.998 g O2)] = 0.005999 mol O2, or 0.006 mol O2</span>
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