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jekas [21]
2 years ago
9

Stoichiometry: Calculate the moles of H2 produced by .042 g Mg in the equation Mg + 2HCl > MgCl2 + H2

Chemistry
1 answer:
Evgesh-ka [11]2 years ago
6 0

Answer:

<em>Mg </em>(<em>s</em>) + 2<em>HCI2 </em>(<em>aq</em>) → <em>MgCI2 </em>(<em>aq</em>) + <em>H2 </em>(<em>g</em>)

I think this is the correct answer I not a 100% sure if it is correct.

Explanation:

Guessing

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The molecular weight of water, H2O, is 18.02 g over mol. How many moles of water are in 24.3 g of water? Express the answer usin
Zolol [24]

Answer:

1.35

Explanation Use the mole formla and multiply and them apply the sig fig rules and you will get the moles of water hope this helps god bless

4 0
3 years ago
!! ANSWER QUICK !!
kari74 [83]

Answer:

I can use a dichotomous key. It helps me classify objects by sorting it out with "yes" and "no" questions.

or

I can use a Punnett Square. It helps me classify what genes the offspring will receive simply by figuring out the recessive and dominant genes as well as the hetzygous and homzygous.

Now give an example of which ever chart you choose by drawing it if that is required. For the Punnett Square label each of the squares Top right Hetzygous, top left dominant, bottom left recessive, bot-tom right homzygous. And for the dichotomous key put a 5-7 length branch showing the animals that have fur, can breathe under water, what cannot or doesn't have those traits. or something similar

Hopefully this helps :)

3 0
3 years ago
Read 2 more answers
(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g
Anarel [89]

Answer:

Part A

 The volume of the gaseous product  is  V = 787L

Part B

The volume of the the engine’s gaseous exhaust is  V_e = 2178 \ L

Explanation:

Part A

From the question we are told that

    The temperature is  T = 350^oC = 350 +273 =623K

     The pressure is  P = 735 \ torr = \frac{735}{760} =  0.967\ atm

     The of  C_8 H_{18} = 100.0g

The chemical equation for this combustion is

               2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}

 The number of moles of  C_8 H_{18} that reacted is mathematically represented as

               n = \frac{mass \ of \  C_8H_{18}  }{Molar \  mass \ of  C_8H_{18} }

The molar mass of  C_8 H_{18} is constant value which is

                  M = 114.23 \ g/mole  

So          n = \frac{100  }{114.23} }

             n = 0.8754 \ moles

The gaseous product in the reaction is CO_2_{(g)} and water vapour

Now from the reaction

    2 moles of C_8 H_{18}  will react with 25 moles of O_2 to give (16 + 18) moles of CO_2_{(g)} and  H_2 O_{(g)}

So

    1 mole of C_8 H_{18} will  react with 12.5 moles of  O_2 to give 17 moles of CO_2_{(g)} and  H_2 O_{(g)}

This implies that

    0.8754 moles of C_8 H_{18} will react with (12.5 * 0.8754 ) moles of O_2 to give  (17 * 0.8754) of CO_2_{(g)} and  H_2 O_{(g)}

So the no of moles of gaseous product is

         N_g = 17 * 0.8754

         N_g = 14.88 \ moles

From the ideal gas law

       PV = N_gRT

making V the subject

        V = \frac{N_gRT}{P}

Where R is the gas constant with a value R = 0.08206 \  L\cdot atm /K \cdot mole

Substituting values

          V = \frac{14.88* 0.08206 *623}{0.967}

          V = 787L

Part B

From the reaction the number of moles of oxygen that reacted is

         N_o = 0.8754 * 12.5

         N_o = 10.94 \ moles

The volume is

      V_o  = \frac{10.94 * 0.08206 *623}{0.967}

      V_o  = 579 \ L

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

         V_e = V_o * \frac{0.79}{0.21}

Substituting values

       V_e = 579 * \frac{0.79}{0.21}

       V_e = 2178 \ L

3 0
3 years ago
In Chromium-53, the "53" represents the
alina1380 [7]

Answer:

The differemt isotopes that differ in atomic mass

Explanation:

4 0
2 years ago
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In a chemical equation, the arrow points towards what?
stepan [7]
The arrow doesn’t “point” to anything really, it means that the reactants form (or change) into the products.
3 0
3 years ago
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