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ExtremeBDS [4]
3 years ago
14

Write the following measurement out of scientific notation. 2.345 x 10^-4 cm.

Chemistry
1 answer:
maw [93]3 years ago
7 0
.0002345 I believe this is correct
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Thanks in advance ❤❤❤❤❤❤​
lesantik [10]

Answer:

A

Divide both sides by 2

m ≤9

4 0
3 years ago
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One of the hydroxybenzoic acids is known by the common name salicylic acid. Its methyl ester, methyl salicylate, occurs in oil o
KIM [24]

Answer:

Please see attachment

Explanation:

Please see attachment

8 0
3 years ago
Each element consists of only one type of atom. <br> a. True<br> b. False
Vladimir79 [104]
A True

Each element consists of one type of atom but, not contains only one atom. It has same type of atoms present in it.
4 0
3 years ago
A 118-ml flask is evacuated and found to have a mass of 97.129 g. when the flask is filled with 768 torr of helium gas at 35 ?c,
Inessa05 [86]
The full question asks to decide whether the gas was a specific gas. That part is missing in your question. You need to decide whether the gas in the flask is pure helium.

To decide it you can find the molar mass of the gas in the flask, using the ideal gas equation pV = nRT, and then compare with the molar mass of the He.

From pV = nRT you can find n, after that using the mass of gass in the flask you use MM = mass/moles.

1) From pV = nRT, n = pV / RT

Data:
V = 118 ml = 0.118 liter
R = 0.082 atm*liter/mol*K
p = 768 torr * 1 atm / 760 torr = 1.0105 atm
T = 35 + 273.15 = 308.15 K

n = 1.015 atm * 0.118 liter / [ 0.082 atm*liter/K*mol  * 308.15K] =0.00472 mol

mass of gas = mass of the fask with the gas - mass of the flasl evacuated = 97.171 g - 97.129 g = 0.042

=> MM =  mass/n = 0.042 / 0.00472 = 8.90 g/mol

Now from a periodic table or a table you get that the molar mass of He is 4g/mol

So the numbers say that this gas is not pure helium , because its molar mass is more than double of the molar mass of helium gas.
7 0
3 years ago
A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions
Mars2501 [29]

Answer:

Qm  = -55.8Kj/mole

Explanation:

NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)

Qm = (mc∆T)water /moles acid

Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)

=> 0.03mole NaOH(aq) + 0.03mole HNO₃(aq)

=> 0.03mole NaNO₃(aq) + 0.03mole H₂O(l)

ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃

= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*

Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C.

4 0
3 years ago
Read 2 more answers
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