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oee [108]
3 years ago
15

I need help on this question

Chemistry
2 answers:
uranmaximum [27]3 years ago
6 0

Answer:

When a front passes over an area, it means a change in the weather. Many fronts cause weather events such as rain, thunderstorms, gusty winds, and tornadoes. At a cold front, there may be dramatic thunderstorms. At a warm front, there may be low stratus clouds. Usually, the skies clear once the front has passed.

Explanation:

Plz make me brainliest :)

Have a nice day

PilotLPTM [1.2K]3 years ago
3 0

Answer:

Email your teacher they will give you the answers try it yourself

Explanation:

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If an object loses mass, what will happen to its weight?
Sophie [7]
The answer is B. The less mass then the less the gravity pulls on it
8 0
3 years ago
Read 2 more answers
When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were
uysha [10]

Answer:

\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}

(b) Mass of H

\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}

(c)Mass of Fe

(i)In 0.4131g of X

\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}

(ii) In 1.2383 g of X

\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}

5 0
3 years ago
Suppose of barium acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of barium cati
ollegr [7]

Explanation:

Let us assume that the given data is as follows.

   mass of barium acetate = 2.19 g

   volume = 150 ml = 0.150 L    (as 1 L = 1000 ml)

   concentration of the aqueous solution = 0.10 M

Therefore, the reaction equation will be as follows.

        Ba(C_{2}H_{3}O_{2})_{2} \rightarrow Ba^{2+} + 2C_{2}H_{3}O^{-}_{2}

Hence, moles of C_{2}H_{3}O^{-}_{2} = 2 \times Ba(C_{2}H_{3}O_{2})_{2}  .......... (1)

As,   No. of moles = \frac{mass}{\text{molar mass}}

Hence, moles of Ba(C_{2}H_{3}O_{2})_{2} will be calculated as follows.                          

     No. of moles = \frac{mass}{\text{molar mass}}  

                          =  \frac{2.19 g}{255.415 g/mol}   (molar mass of Ba(C_{2}H_{3}O_{2})_{2} is 255.415 g/mol)            

                       = 8.57 \times 10^{-3}

    Moles of C_{2}H_{3}O^{-}_{2} = 2 \times 8.57 \times 10^{-3}

                          = 0.01715 mol

Hence, final molarity will be as follows.

              Molarity = \frac{\text{no. of moles}}{volume}

                             = \frac{0.01715 mol}{0.150 L}

                             = 0.114 M

Thus, we can conclude that final molarity of barium cation in the solution is 0.114 M.

5 0
3 years ago
What can electrons hold
-Dominant- [34]
One can only hold 2 electrons which only means shell two can hold 8, and for the first eighteen elements shell three can hold a maximum of eight electrons
5 0
3 years ago
Balancing Al+ Agcl, i have problems with this so help!!!
EleoNora [17]

Answer:

Al  + 3AgCl  →  AlCl₃  + 3Ag

Explanation:

The given equation is:  

      Al  + AgCl  →  

We are to find the product and hence balance the equation. This problem is a simple single replacement reaction.

By virtue of this, Aluminum will displace Ag from the solution:

                       Al  + AgCl  →  AlCl₃  + Ag

    We then balance the equation:

                        Al  + 3AgCl  →  AlCl₃  + 3Ag

                                             

5 0
3 years ago
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