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SVEN [57.7K]
3 years ago
9

What is 93,000,000 in scientific notation?

Chemistry
1 answer:
DanielleElmas [232]3 years ago
5 0
I believe it is 9.3x10^7
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How many grams of the excess reactant remain assuming the reaction goes to completion and that you start with 15.5g of na2s and
tangare [24]
The reaction between Na2S and CuSO4 will give us the balanced chemical reaction of,
                                     Na2S + CUSO4 --> Na2SO4 + CuS
This means that for every 78g of Na2S, there needs to be 159.6 g of CuSO4. The ratio is equal to 0.4887 of Na2S: 1 of CuSO4. Thus, for every 12.1g of CuSO4, we need only 5.91 g of Na2S. Thus, there is an excess of 9.58 g of Na2S. The answer is letter C. 
8 0
3 years ago
Read 2 more answers
I know there are 4.5*10^-9 seconds in 4.5 nanoseconds, but when I enter it into my TI-30xIIs, I get 4.5*10^-10. I enter it as 4.
o-na [289]

Answer:

You are not using properly the function exponential in your calculator

Explanation:

When a number is too big or too small we use scientific notation. This is a number between 1 a 10 multiplied by a power of 10.

When you are writing 4.5*10^-9 you are actually writing 0.0000000045 in scientific notation.

When you enter this in the calculator you have to use the function EXP after the first two numbers.

Steps:    1) Enter 4.5

             2) Enter EXP

             3) Enter minus (-)

             4) Enter 9

8 0
3 years ago
When you climb to the top of a water slide, you are gaining:
Annette [7]

Answer:

C

Explanation:

You climbing the water slide gives you potential energy and your building gravitational eneergy.

3 0
3 years ago
*pls hurry, will give brainlyest or whatever!!*
brilliants [131]

Answer:

all of the above. they all are chemical reactions

4 0
3 years ago
Enter your answer in the provided box. Liquid methanol (CH3OH) can be used as an alternative fuel in pickup and SUV engines. An
andreyandreev [35.5K]

Answer: E=∆H*n= -40.6kj

Explanation:

V(CO) =15L=0.015M³

P=11200Pa

T=85C=358.15K

PV=nRT

n=(112000×0.015)/(8.314×358.15)

n(Co)= 0.564mol

V(Co)= 18.5L = 0.0185m³

P=744torr=98191.84Pa

T= 75C = 388.15k

PV=nRT

n= (99191.84×0.0185)/(8.314×348.15)

n(H2) = 0.634mol

n(CH30H) =1/2n(H2)=1/2×0.634mol

=0.317mol

∆H =∆Hf{CH3OH}-∆Hf(Co)

∆H=-238.6-(-110.5)

∆H = 128.1kj

E=∆H×n=-40.6kj.

3 0
2 years ago
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