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bixtya [17]
3 years ago
11

When na2cro4(aq) and agno3(aq) are mixed, a red colored precipitate forms which is?

Chemistry
1 answer:
Gnoma [55]3 years ago
8 0
As you can see from the given reactants, this is a double-replacement reaction where the metals and nonmetals of both reactants interchange to form two new products. The reaction is written below:

Na₂CrO₄<span>(aq) + 2 AgNO</span>₃<span>(aq)  -->  2 NaNO</span>₃ (aq) + Ag₂CrO₄ (s)

<em>The red colored precipitate is called Silver Chromate, Ag₂CrO₄.</em>


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What volume would 3 moles of hydrogen gas occupy at stp?
Alchen [17]
22.4 L<span>So, if 1 mole occupies 22.4 L, the imediate conclusion is that a bigger number of moles will occupy more than 22.4 L, and a smaller number of moles will occupy less than 22.4 L. In your case, 3 moles of gas will occupy 3 times more volume than 1 mole of gas.</span>
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3 years ago
If 0.25 moles of KBr is dissolved in 0.5 liters of
Len [333]

Answer:

[KBr] = 454.5 m

Explanation:

m is a sort of concentration that indicates the moles of solute which are contianed in 1kg of solvent.

In this case, the moles of solute are 0.25 moles.

Let's determine the mass of solvent in kg.

Density of heavy water, solvent, is 1.1 g/L and our volume is 0.5L.

1.1 g = mass of solvent / 0.5L, according to density.

mass of solvent = 0.5L . 1.1g/L = 0.55 g

We convert the mass to kg → 0.55 g . 1kg /1000g = 5.5×10⁻⁴ kg

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In the diagram, above, marker E is pointing to a __________, which is the “winding” shape that a river can take and its shape is
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Answer:

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Read 2 more answers
A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt
Nadusha1986 [10]

Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt

integrating factor

=e^{\int\limits (1/40) \, dt}

integrating factor =e^{(1/40)t}

multiply on both sides by  =e^{(1/40)t}

dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t

integrate on both sides

\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})

b)

after long period of time means t - > ∞

{t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
6 0
3 years ago
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