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3 years ago

Tell me a funny joke

Chemistry

2 answers:

nasty-shy [4]3 years ago

8 0

Answer:

Explanation:

lol, i tried

dont report, as this is related to math

sp2606 [1]3 years ago

6 0

Answer:

Please mark me brainliest thanks

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When backpacking in the wilderness, hikers often boil water to sterilize it for drinking. Suppose that you are planning a backpa

Pavlova-9 [17]

Answer:

2.104 L fuel

Explanation:

Given that:

Volume of water = 35 L = 35 × 10³ mL

initial temperature of water = 25.0 ° C

The amount of heat needed to boil water at this temperature can be calculated by using the formula:

$q_{boiling} = mc \Delta T$

where

specific heat of water c= 4.18 J/g° C

$q_{boiling} = 35 \times 10^{3} \times \dfrac{1.00 \ g}{1 \ mL} \times 4.18 \ J/g^0 C \times (100 - 25)^0 C$

$q_{boiling} = 10.9725 \times 10^6 \ J$

Also; Assume that the fuel has an average formula of C7 H16 and 15% of the heat generated from combustion goes to heat the water;

thus the heat of combustion can be determined via the expression

$q_{combustion} =- \dfrac{q_{boiling}}{0.15}$

$q_{combustion} =- \dfrac{10.9725 \times 10^6 J}{0.15}$

$q_{combustion} = -7.315 \times 10^{7} \ J$

$q_{combustion} = -7.315 \times 10^{4} \ kJ$

For heptane; the equation for its combustion reaction can be written as:

$C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}$

The standard enthalpies of the products and the reactants are:

$\Delta H _f \ CO_{2(g)} = -393.5 kJ/mol$

$\Delta H _f \ H_2O_{(g)} = -242 kJ/mol$

$\Delta H _f \ C_7H_{16 }_{(g)} = -224.4 kJ/mol$

$\Delta H _f \ O_{2{(g)}} = 0 kJ/mol$

Therefore; the standard enthalpy for this combustion reaction is:

$\Delta H ^0= \sum n_p\Delta H^0_{f(products)}- \sum n_r\Delta H^0_{f(reactants)}$

$\Delta H^0 =( 7 \ mol ( -393.5 \ kJ/mol) + 8 \ mol (-242 \ kJ/mol) -1 \ mol( -224.4 \ kJ/mol) - 11 \ mol (0 \ kJ/mol))$

$\Delta H^0 = (-2754.5 \ \ kJ - 1936 \ \ kJ+224.4 \ \ kJ+0 \ \ kJ)$

$\Delta H^0 = -4466.1 \ kJ$

This simply implies that the amount of heat released from 1 mol of C7H16 = 4466.1 kJ

However the number of moles of fuel required to burn $7.315 \times 10^{4} \ kJ$ heat released is:

$n_{fuel} = \dfrac{q}{\Delta \ H^0}$

$n_{fuel} = \dfrac{-7.315 \times 10^{4} \ kJ}{-4466.1 \ kJ}$

$n_{fuel} = 16.38 \ mol \ of \ C_7 H_{16$

Since number of moles = mass/molar mass

The mass of the fuel is:

$m_{fuel } = 16.38 mol \times 100.198 \ g/mol}$

$m_{fuel } = 1.641 \times 10^{3} \ g$

Given that the density of the fuel is = 0.78 g/mL

and we know that :

density = mass/volume

therefore making volume the subject of the formula in order to determine the volume of the fuel ; we have

volume of the fuel = mass of the fuel / density of the fuel

volume of the fuel = $\dfrac{1.641 \times 10^3 \ g }{0.78 g/mL} \times \dfrac{L}{10^3 \ mL}$

volume of the fuel = 2.104 L fuel

3 0

4 years ago

What is acide, Base and salt

ss7ja [257]

Answer:

In acid – base chemistry, salts are ionic compounds that result from the neutralization reaction of an acid and a base. Basic salts contain the conjugate base of a weak acid, so when they dissolve in water, they react with water to yield a solution with pH greater than 7.0.

Explanation:

Hope this helped!

4 0

3 years ago

Which of the following lists the stages of mitosis in order

AnnyKZ [126]

Anaphase

Metaphase

Prophase

Telophase

Interphase

Prometaphase

Explanation:

4 0

3 years ago

Read 2 more answers

Help I really need to understand this please show how you do it !!

maks197457 [2]

Answer:

HCl

Explanation:

A limiting reactant is the lowest amount of reactant so that the reaction will stop after all that reactant used.

To answer this question, you have to change all reactant units into moles. You have 52g HCl and its molecular mass is 36.46g/mole. The number of HCl in moles will be: 52g/ (36.46g/mole)= 1.43 moles.

In the balanced reaction formula, you can see that you need one mole of HCl and one mole of NaOH for each reaction. Both molecule's coefficient is 1.

If we used up all NaOH, then the number of HCl left will be:

(moles of NaOH used * NaOH coefficient) - (moles of HCl used * HCl coefficient)

(moles of HCl you have * HCl coefficient) - (moles of NaOH used * NaOH coefficient)

(1.43 moles *1 ) - (2.5 moles*1 ) = -1.07 mole

Since the result is minus, then it means we need more HCl and we can't use all NaOH.

If we used up all the HCl, then the number of NaOH left will be:

(moles of NaOH you have* NaOH coefficient) - (moles of HCl used * HCl coefficient)

(2.5 moles*1 )- (1.43 moles *1 )= 1.07 moles

Since the result is plus, then it means we can use all HCl. Then HCl is the limiting reactant

4 0

3 years ago

Can someone help me Balance chemical equations please

horrorfan [7]

2NaClO3 = 2NaCl + 3O2
Hopes this helps <33

4 0

3 years ago