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sergeinik [125]
3 years ago
6

Does anybody know this also explain

Chemistry
1 answer:
Tresset [83]3 years ago
8 0

Answer:

brainliest answer

I believe this is correct

Explanation:

CaBr2+Na2Co3》CaCo3+2NaBr

The double-replacement reaction generally takes the form of AB + CD → AD + CB where A and C are positively-charged cations, while B and D are negatively-charged anions.

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A 225. mL sample 3.5M solution of KBr is diluted to a final volume of 750. mL. What is the final concentration of this solution?
Allushta [10]
C= comcentration
V= volume
C1*V1=C2*V2
3.5M*225.mL = xM*750 mL
x=(3.5*225/750) =1.05M≈1.1M
7 0
4 years ago
Consider the reaction: 2 SO2(g)+O2(g)→2 SO3(g) If 285.5 mL of SO2 reacts with 158.9 mL of O2 (both measured at 315 K and 50.0 mm
Tanzania [10]

Answer:

a. Oxygen is the limiting reagent. n_{SO_3}^{Theoretical}=8.096x10^{-4}mol SO_3

b. Y=58.9%

Explanation:

Hello,

a. Limiting reagent and sulfur trioxide's theoretical yield.

At first, we must compute the involved moles for both sulfur dioxide's and oxygen's as follows, considering the volumes in liters and the pressure in atm of 50.0mmHg*1atm/760mmHg=0.0658atm:

n_{SO_2}=\frac{PV}{RT}=\frac{0.0658atm*0.2855L}{0.082\frac{atm*L}{mol*K}*315K} =7.273x10^{-3}molSO_2 \\n_{O_2}=\frac{PV}{RT}=\frac{0.0658atm*0.1589L}{0.082\frac{atm*L}{mol*K}*315K} =4.048x10^{-4}molO_2

Afterwards, by considering the properly balanced chemical reaction:

2SO_2(g)+O_2(g)-->2SO_3

We compute the oxygen's moles that completely reacts with the previously computed 7.273x10^{-3} moles of SO_2 as follows:

7.273x10^{-3}molSO_2*\frac{1molO_2}{2molSO_2} =3.6365x10^{-3}molO_2

That result let us know that the oxygen is the limiting reagent since just 4.048x10^{-4} moles are available in comparison with the 3.6365x10^{-3} moles that completely would react with 7.273x10^{-3} moles of SO_2.

Now, to compute the theoretical yield of sulfur trioxide, we apply the following stoichiometric relationship:

n_{SO_3}^{Theoretical}=4.048x10^{-4}molO_2*\frac{2molSO_3}{1molO_2} =8.096x10^{-4}mol SO_3

b. Percent yield.

At first, we must compute the collected (real) moles of sulfur trioxide:

n_{SO_3}^{real}=\frac{PV}{RT}=\frac{0.0658atm*0.1872L}{0.082\frac{atm*L}{mol*K}*315K} =4.769x10^{-4}molSO_3

Finally, we compute the percent yield:

Y=\frac{n_{SO_3}^{real}}{n_{SO_3}^{Theoretical}} *100%

Y=\frac{4.769x10^{-4}mol SO_3}{8.096x10^{-4}mol SO_3} *100%

Y=58.9%

Best regards.

7 0
3 years ago
Read 2 more answers
Please help ASAP
aev [14]

Answer:

1st one is correct, Igneous rocks are formed when molten rock cools and hardens.

5 0
3 years ago
Read 2 more answers
Ten moles of hydrogen are allowed to react with 6 moles of oxygen. How much water will be
Anastaziya [24]

Answer:

10 moles of water are produced,

Explanation:

Given data:

Moles of H = 10 mol

Moles of O = 6 mol

Water obtained = ?

Solution:

Balance chemical equation:

2H₂ + O₂    →    2H₂O

Now we will compare the moles of H₂ and O₂ with water from balance chemical equation.

                            H₂             :          H₂O

                             2              :            2

                             10             :           10

                            O₂             :          H₂O

                             1              :            2

                             6             :           2×6 = 12

Number of moles of water produced by hydrogen are less so hydrogen will be limiting reactant and it will limit the yield of water thus, 10 moles of water are produced.

                                 

3 0
3 years ago
Animal Reproduction Interactivity
Leviafan [203]

<u>Step</u><u> </u><u>1</u>

Jellyfish do not give birth to their young because they expel undeveloped eggs in the form of zygote or embryo from their bodies and deposit them on the ocean floor before moving on. They don't even care about the eggs that have been laid.

<u>Step</u><u> </u><u>2</u>

Jellyfish have two distinct body types during their lives: medusa and polyps. Polyps reproduce by budding while medusa reproduce as by spawning eggs and sperm.

Jellyfish can reproduce in the adult, or medusa, stage by releasing sperm and eggs into the water, forming a planula. Polyps clone themselves and bud, or strobilate, into ephyra, a different stage of jellyfish life. The adult medusa jellyfish develops from his form.

Only a few jellyfish species obtain sperm through their mouths in order to fertilise eggs within their bodies, but the majority of jellyfish simply release sperm or eggs into the water. They will do this once a day, normally at dawn or dusk, if conditions are favorable.

4 0
3 years ago
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