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3 years ago

How many sig figs does 0.007 have

Chemistry

1 answer:

jeka57 [31]3 years ago

5 0

Answer:

Explanation:

zeros are not counted when they come before a natural number

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Question 19 Unsaved

aliina [53]

It described a nucleus surrounded by a large volume of space.

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3 years ago

Balancing out Fe +h2O fe3O4 +H2

Katyanochek1 [597]

3Fe + 4H2O (yields) Fe3O4 + 4H2

4 0

3 years ago

Consider the following reaction:C2H4(g) + F2(g) -----------> C2H4F2(g) Delta H = -549 kJEstimate the carbon-fluorine bond ene

frutty [35]

Answer:

Bond energy of carbon-fluorine bond is 485 kJ/mol

Explanation:

Enthalpy change for a reaction, is given as:

$\Delta H_{rxn}=\sum [n_{i}\times (E_{bond})_{i}]-\sum [n_{j}\times (E_{bond})_{j}]$

Where $(E_{bond})_{i}$ and $(E_{bond})_{j}$ represents average bond energy in breaking "i" th bond and forming "j" th bond respectively. $n_{i}$ and $n_{j}$ are number of moles of bond break and form respectively.

In this reaction, one mol of C=C, four moles of C-H and one mol of F-F bonds are broken. One mol of C-C bond, four moles of C-H bonds and two moles of C-F bonds are formed

So, $-549kJ=(1mol\times 614kJ/mo)+(4mol\times E_{C-H})+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(4mol\times E_{C-H})-(2mol\times E_{C-F})$

or, $-549kJ=(1mol\times 614kJ/mo)+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(2mol\times E_{C-F})$

or, $E_{C-F}=485kJ/mol$

So bond energy of carbon-fluorine bond is 485 kJ/mol

8 0

3 years ago

Please help!! will give brainliest if u tell me how

Alona [7]

I am pretty sure it’s A

8 0

3 years ago

125 ml of nitrogen gas is collected at 70.0 degrees Celsius. The pressure

dybincka [34]

Answer: Volume of the gas at STP is 22.53 L.

Explanation:

Given : Volume = 125 mL (as 1 mL = 0.001 L) = 0.125 L

Temperature = $70^{o}C = (70 + 273) K = 343 K$

Pressure = $125 kPa = 125 kPa \times \frac{0.01 atm}{1 kPa} = 1.25 atm$

According to the ideal gas equation, the volume of given nitrogen gas is calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$1.25 atm \times V = 1 mol \times 0.0821 L atm/mol K \times 343 K\\V = \frac{1 mol \times 0.0821 L atm/mol K \times 343 K}{1.25 atm}\\= \frac{28.1603}{1.25} L\\= 22.53 L$

Hence, volume of the gas at STP is 22.53 L.

5 0

3 years ago