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OleMash [197]
3 years ago
6

A big ______ shows up on a weather map as really close/tight isobars & in weather as fast, strong winds

Chemistry
2 answers:
Shtirlitz [24]3 years ago
5 0
The Answer is Precipitation
Ksju [112]3 years ago
3 0

Answer:

A BIG DI*CK

Explanation:

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for the reaction shown compute the theoretical yield of product in moles each of the initial quantities of reactants. 2 Mn(s)+3
Y_Kistochka [10]

Answer:

2 mole MnO₂

Explanation:

2Mn(s) + 2O₂(g) => 2MnO₂(s)

4 0
4 years ago
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Which of the following is NOT a characteristic of metals?
lesantik [10]
I can’t see the question :/
5 0
3 years ago
Which term describes the composition of matter shown in figure on the left? element compound mixture phase​
dangina [55]

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bb

Explanation:

6 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
Assume a hydrogen atom has an electron in the n-level corresponding to the month of your birthday (jan = 1, feb = 2, etc.) calcu
luda_lava [24]
My Birthday is in April, So, electron of H atom is in n=4( pfund series).

Ionisation Energy= +13.6(z^2/n^2)
= +13.6(1/16)
= +0.85 eV
6 0
3 years ago
Read 2 more answers
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