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3 years ago

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Hey can someone pls help me answer these questions? It should be easy maybe. It’s just science. The question just won’t to know

where you would put the arrow above to show the motion of the plates. Where would I put it?

1 answer:

miskamm [114]3 years ago

6 0

Hehehehwgwgw. Be the hardest thing ever for a long day and I have a windows

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Mn2(CO3)3 what is the cation and anion in this compound?

brilliants [131]

Mn+2 is cation and CO3 is anion
hope it help

7 0

3 years ago

Read 2 more answers

Temperate deciduous trees lose their leaves in Fall. Explain why trees in temperate rainforest and tropical rainforest don’t los

andrey2020 [161]

Answer:

So trees in temperate don't lose their leaves because the weather events aren't harsh enough.

Trees in tropical rainforest don't lose their leaves because they are a different type of tree known as evergreens that are green all year round.

Explanation:

Ok so first we'll define some things

Deciduous Trees= Trees that lose all of their leaves for part of the year.

Trees shed their leaves trees to try and survive harsh weather events.

Temperate deciduous trees lose their leaves in fall to better survive the winter conditions of extreme cold and reduced daylight.

Temperate rainforests = An area that doesn't experience extremely cold or extremely hot temperatures or what we would call harsh weather events.

Broad-leaved trees in tropical rainforests are known evergreen, they are known as this as they are green all year round.

6 0

3 years ago

How much heat is released when you condense 93.9 g of water vapor? <br><br> Show work please ❤️

sammy [17]

<u>Answer:</u>

211.9 J

<u>Explanation:</u>

The molecules of water release heat during the transition of water vapor to liquid water, but the temperature of the water does not change with it.

The amount of heat released can be represented by the formula:

$Q=mL_e$

where $Q$ = heat energy, $m$ = mass of water and $L$ = latent heat of evaporation.

The latent heat of evaporation for water is $L=2257 kJ/kg$ and the mass of the water is $m=93.9 g=0.0939 kg$ .

The amount of heat released in this process is:

$Q=mL_e = (0.0939kg)(2257 kJ/kg)=$ 211.9 J

7 0

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.

5 0

3 years ago

AfilCa [17]

Answer:

the fourth one I think....m

Explanation:

tfffffdd

7 0

3 years ago