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Monica [59]
3 years ago
8

so i am doing a science project and i am doing it on mold i put water on the bread and cheese and left one in the dark and one i

n the light FOR A TWO WEEKS it isnt working why not
Chemistry
1 answer:
ivann1987 [24]3 years ago
3 0

Answer:

because the bread and cheese is absorbing the water whereas if you leave it dry in a moist humid place mold will start to appear

Explanation:

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Answer: I think G.

Explanation:

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The distance from the sun to earth would be ______.
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C- more than one light year or B-exactly one light year
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What would happen if the aspects that allow to maintain the characteristics of the states were altered?​
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3 0
3 years ago
The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
MAVERICK [17]

Answer:

Approximately 2.46\; \rm mol.

Explanation:

Make use of the molar mass data (M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}) to calculate the number of moles of molecules in that 64.0\; \rm g of \rm C_2H_2:

\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}.

Make sure that the equation for this reaction is balanced.

Coefficient of \rm C_2H_2 in this equation: 2.

Coefficient of \rm H_2O in this equation: 2.

In other words, for every two moles of \rm C_2H_2 that this reaction consumes, two moles of \rm H_2O would be produced.

Equivalently, for every mole of \rm C_2H_2 that this reaction consumes, one mole of \rm H_2O would be produced.

Hence the ratio: \displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1.

Apply this ratio to find the number of moles of \rm H_2O that this reaction would have produced:

\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}.

3 0
3 years ago
If the atmospheric pressure is 0.975 atm what is the pressure of the enclosed gas
Nuetrik [128]
Missing in your question:
Picture (1)
when its an open- tube manometer and the h = 52 cm. 
when the pressure of the atmosphere is equal the pressure of the gas plus the pressure from the mercury column 52 Cm so, we can get the pressure of the gas from this formula:
P(atm) = P(gas) + height (Hg)
∴P(gas) = P(atm) - height (Hg)
              = 0.975 - (520/760) 
              = 0.29 atm
Note: I have divided 520 mm Hg by 760 to convert it to atm
Picture (2)
The pressure of the gas is the pressure experts by the column of mercury and when we have the Height (Hg)= 67mm 
So the pressure of the gas =P(atm) + Height (Hg)
                                             =  0.975 + (67/ 760) = 1.06 atm
Picture (3) 
As the tube is closed SO here the pressure of the gas is equal the height of the mercury column, and when we have the height (Hg) = 103 mm. so, we can get the P(gas) from this formula:
P(gas) = Height(Hg)
           = (103/760) = 0.136 atm

6 0
2 years ago
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