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2 years ago

Hey what's 10 & 11 y'all im dumb

Chemistry

1 answer:

lesya [120]2 years ago

3 0

Answer:

10 is 2.89g/ml

11 in attachment

Explanation:

10. D=M/V

M=250G-120G=130G

V=75ML-30ML=45ML

D=130/45=2.89g/ml

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I'm not 100% sure of the answer. Explanations are highly appreciated.

PolarNik [594]

Answer:

The correct answer is

$\Delta H=2(1072)+498 -4(799)$

Explanation:

Enthalpy of reaction :

It is the amount of energy released/absorbed when one mole of the substance is formed from the reactant at a constant pressure.

The enthalpy of a reaction can be calculated using :

$\Delta H=\Delta H_{reactants}-\Delta H_{products}$

$2CO+O_{2}\rightarrow 2CO_{2}$

$\Delta H_{reactants}=2(C\equiv O)+O=O$

$\Delta H_{reactants}=2(1072)+498$

$H_{products}=2(2\times C=O)$

$H_{products}=4\times 799$

$\Delta H=\Delta H_{reactants}-\Delta H_{products}$

$\Delta H=2(1072)+498 -4(799)$

$units = \frac{kJ}{mole}$

Please note that :

The carbon monoxide , CO should be taken as C triple bond O. Not C=O .

So , the bond energy =1072 is used

$\Delta H=2(1072)+498 -4(799)$

8 0

3 years ago

What is the specific latent heat of fusion of ice if it takes 863 kJ to convert 4.6 kg of ice into water at 0 C?

allochka39001 [22]

Answer:

$1^{f} =187.7 \frac{J}{kg}$

Explanation:

SO in order to calculate the specific latent heat of fusion, you need to remember the formula:

$1^{f} =\frac{Q}{m}$

Where $1^{f}$ representes the specific latent heart of fusion.

$Q$ represents the heat energy added, usually represented in kJ

$m$ represents the mass of the object, in kg.

Now that we have our formula we just have to put our values into the formula:

$1^{f} =\frac{Q}{m}$

$1^{f} =\frac{863kJ}{4.6kg}$

$1^{f} =187.7 \frac{J}{kg}$

SO our answer would be $1^{f} =187.7 \frac{J}{kg}$

7 0

3 years ago

Consider an electron with charge âˆ’eâˆ’e and mass mmm orbiting in a circle around a hydrogen nucleus (a single proton) with cha

PtichkaEL [24]

Answer:

Explanation:

The net force on electron is electrostatic force between electron and proton in the nucleus .

Fc = $\frac{1}{4\pi\epsilon} \times \frac{e\times e}{r^2}$

This provides the centripetal force for the circular path of electron around the nucleus .

Centripetal force required = $\frac{m\times v^2}{r}$

$\frac{m\times v^2}{r}=\frac{1}{4\pi\epsilon} \times \frac{e\times e}{r^2}$

$v^2=\frac{e^2}{4\pi \epsilon m r}$

$v=(\frac{e^2}{4\pi \epsilon m r})^{\frac{1}{2} }$

5 0

2 years ago

Oxidation number ofAl2S12O7

Savatey [412]

Answer:

-2? if wrong im sorry please tell me what I did that was incorrect ty Explanation:

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2 years ago

Please help i will mark brainliest which helps bring up youre rank

Elis [28]

Answer:

push or pull

Explanation:

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3 years ago