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Yanka [14]
3 years ago
9

When the following equation is balanced what is the coefficient for CaCI2? Ca (s) + 2HCI (aq) —> CaCI2 (aq) + H 2(g)

Chemistry
2 answers:
Allisa [31]3 years ago
7 0

Hello!

The coefficients of any chemical compound are located in the front.

We are asked to find the coefficient of CaCl2 when the equation is balanced. Before finding the coefficient, we should determine if the equation is balanced or not.

Right side: 1 Ca, 2 H, 2 Cl

Left side: 1 Ca, 2 H, 2 Cl → This equation is balanced!

Since the compound CaCl2 has no number in from of the compound, the coefficient has to 1.

Therefore, the coefficient for CaCl2 is 1.

artcher [175]3 years ago
3 0

The coefficients of any chemical compound are located in the front.


We are asked to find the coefficient of CaCl2 when the equation is balanced. Before finding the coefficient, we should determine if the equation is balanced or not.


Right side: 1 Ca, 2 H, 2 Cl


Left side: 1 Ca, 2 H, 2 Cl → This equation is balanced!


Since the compound CaCl2 has no number in from of the compound, the coefficient has to 1.


Therefore, the coefficient for CaCl2 is 1.

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Answer:

Curvature

Explanation:

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100 POINTS! Final Honor Activity Question
castortr0y [4]

The change in temperature had the greatest effect at changing the volume of the balloon.

<h3>What are the gas laws?</h3>

The gas laws are used to describe the parameters that has to do with gases.

Given that;

P1 = 98.5 kPa

T1 = 18oC or 291 K

V1 =  74.0 dm3

P2 =  7.0 kPa

V2 = ?

T2 = 18oC or 291 K

P1V1/T1 = P2V2/T2

P1V1T2 =P2V2T1

V2= P1V1T2/P2T1

V2 =  98.5 kPa *  74.0 dm3 * 291 K/ 7.0 kPa * 291 K

V2 = 1041.3 dm3

When;

V1 = 1041.3 dm3

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T2 = 80oC or 353 K

V1/T1 = V2/T2

V1T2 = V2T1

V2 = V1T2/T1

V2 = 1041.3 dm3 * 353 K/291 K

V2 = 1263 dm3

The change in temperature had the greatest effect at changing the volume of the balloon.

Given that

V1 =  100 cm^3

T1 = 273 K

P1 = 1.01 * 10^5 Pa

V2 = ?

P2 =  3.00 x 10^-4 Pa

T2 = -180oC or 255 K

V2= P1V1T2/P2T1

V2 =  1.01 * 10^5 Pa * 100 cm^3 * 255 K / 3.00 x 10^-4 Pa * 273 K

V2 = 3.14 * 10^10 cm^3

Learn more about gas laws:brainly.com/question/12669509

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7 0
2 years ago
7. Diethyl ether burns in air according to the following equation. C4H10O(l) + 6O2 (g) → 4CO2 (g) + 5H2O(l) If 7.15 L of CO2 is
iren [92.7K]

From the stoichiometry of the combustion reaction, we can see that 7.4 L of oxygen is consumed.

<h3>What is combustion?</h3>

Combustion is a reaction in which a substance is burnt in oxygen. The equation of the reaction is; C4H10O(l) + 6O2 (g) → 4CO2 (g) + 5H2O(l)

We can obtain the number of moles of CO2 from;

PV = nRT

n = 1.02 atm * 7.15 L/0.082 atm LK-1mol-1 * (125 + 273) K

n = 7.29 /32.6

n = 0.22 moles

If 6 moles of oxygen produces 4 moles of CO2

x moles of oxygen produces 0.22 moles of CO2

x = 0.33  moles

1 mole of oxygen occupies 22.4 L

0.33 moles of oxygen occupies 0.33 moles *  22.4 L/ 1 mole

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6 0
3 years ago
Which of the following are weak electrolytes in water
anzhelika [568]
Not strong base and acid, not dissolved or not aqueous.
4 0
3 years ago
Read 2 more answers
Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.
Ad libitum [116K]

Explanation:

The given data is as follows.

    Concentration = 0.1 mol/dm^{3}

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

                             = 6.022 \times 10^{25} ions/m^{3}

               T = 30^{o}C = (30 + 273) K = 303 K

Formula for electric double layer thickness (\lambda_{D}) is as follows.

            \lambda_{D} = \frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}

where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

                          = 1 nm (approx)

Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

7 0
3 years ago
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