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3 years ago

8

I have a balanced equation. I can convert moles of one substance to moles of another substance using the _____ ratio as a unit f

actor

1 answer:

Olin [163]3 years ago

7 0

Answer: sorry can't help ya

Explanation:

Sorry it's just I don't really know how to answer this question myself wish you the best of luck ;3

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Paha777 [63]

The answer should be d

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3 years ago

What is the modren system of classificationis based on

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In modern biology, there are three approaches to classifying organisms: systematics, cladistics and molecular evolutionary taxonomy. They are all based on organisms' relation to each other, but use different indicators to assign the degree of relationship

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3 years ago

Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of

Brut [27]

<u>Answer:</u> The smallest temperature change is shown by water.

<u>Explanation:</u>

To calculate the heat absorbed or released, we use the equation:

$q=mC\times \Delta T$ ......(1)

where,

q = heat absorbed = 200.0 J

m = mass of the substance

C = specific heat of substance

$\Delta T$ = change in temperature

As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

<u>Option a:</u> 50.0 g Fe, $C_{Fe}=0.449J/g^oC$

We are given:

$m=50.0g\\C_{Fe}=0.449J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC$

Change in temperature = 8.99°C

<u>Option b:</u> 50.0 g water, $C_{water}=4.18J/g^oC$

We are given:

$m=50.0g\\C_{water}=4.18J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC$

Change in temperature = 0.96°C

<u>Option b:</u> 25.0 g Pb, $C_{Pb}=0.128J/g^oC$

We are given:

$m=50.0g\\C_{Pb}=0.128J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.128J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.128}=62.5^oC$

Change in temperature = 62.5°C

<u>Option d:</u> 25.0 g Ag, $C_{Ag}=0.235J/g^oC$

We are given:

$m=25.0g\\C_{Ag}=0.235J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC$

Change in temperature = 34.04°C

<u>Option e:</u> 25.0 g granite, $C_{granite}=0.79J/g^oC$

We are given:

$m=25.0g\\C_{Fe}=0.79J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.79J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.79}=10.13^oC$

Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

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7) 3 H₂SO₄ + 2 Al(OH)₃ → Al₂(SO₄)₃ + 6 H₂O

8) Cd(NO₃)₂ + Na₂S → CdS + 2 NaNO₃

9) Cr₂(SO₄)₃ + 3 (NH₄)₂CO₃ → Cr₂(CO₃)₃ + 3 (NH₄)₂SO₄

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A negative membrane potential was recorded when the tip of the microelectrode was

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Answer; both inside the cell body and inside the axon.

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