What type of bond is present between two atoms of tungsten in a light bulb filament?

PLEASE HELP WILL GIVE 20PTS AND WILL MARK BRAINLIEST FOR RIGHT ANSWER!!!

Marysya12 [62]

Answer:

C) 0.457

Explanation:

The ratio between O2 and H2O is 1:2 according to the balanced equation. You can find how many moles is O2 by : 5.12/22.4 = 0.22857 ( 1 mole = 22.4 litters)

Moles of H2O will be 0.22857 * 2 = 0.457142.

Therefore answer C)

7 0

3 years ago

How did modern scientists experiment with Van Goghs paints to determine their chemical reactions?

Vlada [557]

Answer:

the would test random expirements

7 0

2 years ago

If 45.0 mL of ethanol (density=0.789 g/mL) initially at 9.0 C is mixed with 45.0 mL of water (density=1.0 g/mL) initially at 28.

Klio2033 [76]

Answer : The final temperature of the mixture is $22.7^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

$(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ethanol = $2.3J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ethanol

$m_2$ = mass of water

$\rho_1$ = density of ethanol = 0.789 g/mL

$\rho_2$ = density of water = 1.0 g/mL

$V_1$ = volume of ethanol = 45.0 mL

$V_2$ = volume of water = 45.0 mL

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of ethanol = $9.0^oC$

$T_2$ = initial temperature of water = $28.6^oC$

Now put all the given values in the above formula, we get

$(0.789g/mL\times 45.0mL)\times (2.3J/g^oC)\times (T_f-9.0)^oC=-(1.0g/mL\times 45.0mL)\times 4.18J/g^oC\times (T_f-28.6)^oC$

$T_f=22.7^oC$

Therefore, the final temperature of the mixture is $22.7^oC$

4 0

3 years ago

Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2

PIT_PIT [208]

Answer: The standard heat for the given reaction is -138.82 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ$

Hence, the standard heat for the given reaction is -138.82 kJ

3 0

3 years ago

On a potential energy diagram for the following processes, which of the following has an increase in entropy?

daser333 [38]

Hello,

I believe that your answer would be B. water freezes
Hope this helps

7 0

3 years ago

Read 2 more answers