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Dmitriy789 [7]
2 years ago
10

What is the correct formula that would result from the combination of the two ionic species?

Chemistry
1 answer:
Soloha48 [4]2 years ago
6 0

Answer:

CsCl

Explanation:

The ionic species are:

           Cs⁺   and Cl⁻

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Calculate the number of grams of solute in 814.2mL of 0.227 M calcium acetate
kiruha [24]

Answer:

Mass = 29.23 g

Explanation:

Given data:

Volume of solution = 814.2 mL 814.2/1000 = 0.8142 L)

Molarity of solution = 0.227 M

Mass of solute in gram = ?

Solution:

Molarity = number of moles / volume in L

By putting values,

0.227 M = number of moles / 0.8142 L

Number of moles = 0.227 M × 0.8142 L

Number of moles = 0.184 mol

Mass in gram:

Mass = number of moles × molar mass

Molar mass of calcium acetate = 158.17 g/mol

Mass = 0.184 mol × 158.17 g/mol

Mass = 29.23 g

6 0
3 years ago
The volume of oxygen, collected over water, is 185 mL at 25 degrees Celsius and 600 torr. calculate the dry volume of the oxygen
ivolga24 [154]

Answer:

0.1593 L.

Explanation:

  • We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If n and P are constant, and have two different values of V and T:

<em>P₁V₁T₂ = P₂V₂T₁</em>

<em></em>

P₁ = 600 torr/760 = 0.789 atm, V₁ = 185.0 mL = 0.185 L, T₁ = 25.0°C + 273 = 298.0 K.

P₂ (at STP) = 1.0 atm, V₂ = ??? L, T₂ (at STP = 0.0°C) = 0.0°C + 273 = 273.0 K.

<em>∴ V₂ = P₁V₁T₂/P₂T₁</em> = (0.789 atm)(0.185 mL)(298.0 K)/(1.0 atm)(273.0 K) = <em>0.1593 L.</em>

4 0
3 years ago
The day to day condition of atmosphere is known as
vovikov84 [41]

Answer:

d) atmosphere pressure

5 0
3 years ago
Read 2 more answers
The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (
Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^{x}[B]^{y}[C]^{z}

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}

\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}

3 = 3^{x} \\\\x =1

Order w.r.t B : Use trials 2 and 5

\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}

\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}

4 = 2^{y} \\\\y =2

Order w.r.t C : Use trials 1 and 2

\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}

1 = (0.5)^{z}

z = 1

Therefore, order w.r.t C = 1

8 0
3 years ago
Determine the temperature change when 150. G block of gold is supplied with 1.00 • 10 ^3 J of heat
IceJOKER [234]

Answer:

∇T = 51.68°C

Explanation:

Mass = 150g

Heat Energy (Q) = 1.0*10³J

Change in temperature ∇T = ?

Q = mc∇T

Q = heat energy

M = mass

C = specific heat capacity of the gold = 0.129j/g°C

∇T = change in temperature

Q = Mc∇T

1.0*10³ = 150 * 0.129 * ∇T

1000 = 19.35∇T

Solve for ∇T

∇T = 1000 / 19.35

∇T = 51.679°C = 51.68°C

The change in temperature of gold was 51.68°C

8 0
3 years ago
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