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3 years ago

2 Which two substances cause iron to rust?

Chemistry

1 answer:

guapka [62]3 years ago

8 0

Answer:

water and oxygen

Explanation:

this is because

iron + water + oxygen ----> hydrated iron (III) oxide (which is rust in lame terms )

hope this helps

please mark it brainliest

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Suppose 0.09886 M KOH is titrated into 15.00 mL H2SO4 of unknown concentration until the equivalence point is reached. It takes

marta [7]

Answer:

2 Mol base : 1 Mol acid

Explanation:

balance equation

5 0

3 years ago

Read 2 more answers

how many grams of F2 are needed to react with 3.50 grams of Cl2 Equation needed for question- Cl2+3F2-->2ClF3 please explain

pshichka [43]

Answer:

5.62 g of F2

Explanation:

We have to start with the chemical reaction:

$Cl_2~+~3F_2~-->~2ClF_3$

We have a balanced reaction, so we can continue with the mol calculation. For this, we need to know the molar mass of $Cl_2$ (70.906 g/mol), so:

$3.5~
g~Cl_2
\frac{1~mol~Cl_2}{70.906~g~Cl_2}=0.049~mol~Cl_2$

Now, with the molar ratio between $Cl_2$ and $F_2$ we can convert from moles of $Cl_2$ and $F_2$ (1:3), so:

$0.049~mol~Cl_2\frac{3~mol~F_2}{1~mol~Cl_2}=0.148~mol~F_2$

Finally, with the molar mass of $F_2$ we can calculate the gram of $F_2$ (37.99 g/mol), so:

$0.148~mol~F_2\frac{37.99~g~F_2}{1~mol~F_2}=5.62~g~F_2$

I hope it helps!

8 0

3 years ago

Ignoring sign which transition is associated with the greatest energy change?

scoray [572]

Well u said ignoring not me

3 0

3 years ago

The energy of any one-electron species in its nth state (n = principal quantum number) is given by E = –BZ2 /n2 where Z is the c

Ivahew [28]

Explanation:

(a) The given data is as follows.

B = $2.180 \times 10^{-18} J$

Z = 4 for Be

Now, for the first excited state $n_{f}$ = 2; and $n_{i} = \infinity$ if it is ionized.

Therefore, ionization energy will be calculated as follows.

I.E = $\frac{-Bz^{2}}{\infinity^{2}} - (\frac{-2.180 \times 10^{-18} J /times (4)^{2}}{(2)^{2}})$

= $8.72 \times 10^{-18} J$

Converting this energy into kJ/mol as follows.

$8.72 \times 10^{-18} J \times 6.02 \times 10^{23} mol$

= 5249 kJ/mol

Therefore, the ionization energy of the $Be^{3+}$ ion in its first excited state in kilojoules per mole is 5249 kJ/mol.

(b) Change in ionization energy is as follows.

$\Delta E = -Bz^{2}(\frac{1}{(4)^{2}} - {1}{(2)^{2}}) = \frac{hc}{\lambda}$

$\frac{hc}{\lambda} = 0.1875 \times 2.180 \times 10^{-18} J \times (4)^{2}$

$\lambda = \frac{6.626 \times 10^{-34} \times 2.998 \times 10^{8} m/s}{0.1875 \times 2.180 \times 10^{-18} J \times 16}$

= $303.7 \times 10^{-10} m$

or, = $303.7^{o}A$

Therefore, wavelength of light given off from the $Be^{3+}$ ion by electrons dropping from the fourth (n = 4) to the second (n = 2) energy levels $303.7^{o}A$ .

5 0

3 years ago

When nitrogen gas (N2) reacts with hydrogen gas(H) ammonia gas (NH3) is formed. How many grams of hydrogen gas are required to r

kupik [55]

Mass of Hydrogen gas required to react : 0.936 g

<h3>Further explanation</h3>

Reaction on Nitrogen gas and Hydrogen gas to produce Ammonia gas

N₂ (g) + 3 H₂ (g) ⇒ 2 NH₃ (g)

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol

so mol Nitrogen for 3.5 L at STP :

$\tt mol=\dfrac{3.5~L}{22.4~L}=0.156$