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andrew11 [14]
2 years ago
15

The difference of a number X and 1/2(how to write it out?)

Chemistry
1 answer:
KatRina [158]2 years ago
6 0
X - (1/2)

I'm pretty sure that's the answer
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I don't unerstand what I'm supposed to do for these
shepuryov [24]
You just need to multiply the terms and transform to kj by dividing by 1000
1) 10.4kj
2) 14.5 kj 
3) 44.8 kj
4) 1.04 kj

and there you go.
8 0
3 years ago
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a cylinder rod formed from silicon is 46.0 cm long and has a mass of 3.00 kg. the density of silicon is 2.33 g/cm³. what is the
Gnesinka [82]
d=\frac{m}{V}\\\\
V=\pi r^{2}H\\d=\frac{m}{\pi r^{2}H} \ \ \ |*\pi r^{2}H\\\\
d \pi r^{2}H=m \ \ \ |:d \pi H\\\\
r^{2}=\frac{m}{d \pi H}\\\\
r=\sqrt{\frac{m}{d \pi H}}\\\\
2r=d\\\\
d=2\sqrt{\frac{m}{d \pi H}}

H=43cm\\
d=2.33\frac{g}{cm^{3}}\\
m=3kg=3000g\\\\
d=2\sqrt{\frac{m}{d \pi H}}=2*\sqrt{\frac{3000g}{2.33\frac{g}{cm^{3}} * \pi * 43cm}}=2*\sqrt{\frac{3000}{100,19\pi \frac{1}{cm^{2}}}}\approx6,17cm
7 0
3 years ago
A chemical reaction was carried out by mixing 25 g of pure CaCO3 and 0.75 mole of pure HCl to give CaCl2, H2O and CO2. a. Which
Naily [24]

hola, esta pregunta es bastante difícil pero está bien, no lo sé, lo siento :) :)

4 0
3 years ago
What would be the number of oxygen atoms in 5.00 moles of co2?
malfutka [58]
Well theres 5 so times 2 equals 10

6 0
3 years ago
A 0.04358 g sample of gas occupies 10.0-ml at 292.0 k and 1.10 atm. upon further analysis, the compound is found to be 25.305% c
kirill [66]
<span>Answer: A 0.04403 g sample of gas occupies 10.0-mL at 289.0 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C and 74.695% Cl. What is the molecular formula of the compound? --------------------------------------... Seems like I did a problem very similar to this--this must be the "B" test. But the halogen was different. 25.305% C/12 = 2.108 74.695% Cl/35.5 = 2.104 So the empirical formula would be CH. However, there are many compounds which fit this bill, so we have to use the gas data. (And I made, in the previous problem, the simplifying assumption that 289C and 1.10 atm would offset each other, so I'll do that, too.) 0.044 grams/10 ml = x/22.4 liters 0.044g/0.010 liters = x/22.4 liters 22.4 liters/0.010 liters = 2240 (ratio) 2240 x .044 = 98.56 (actual atomic weight) CCl = 35.5+12 or 47.5, so two of those is 95 grams/mole. This is sufficiient to distinguish C2CL2, (dichloroacetylene) from C6CL6 (hexachlorobenzene) which would mass 3 times as much.</span>
3 0
3 years ago
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