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Ray Of Light [21]
3 years ago
8

How many grams of oxygen are needed to react completely with 200.0 g of ammonia, nh3? 4nh3(g) + 5o2(g) ®4no(g) + 6h2o(g)

Chemistry
1 answer:
VARVARA [1.3K]3 years ago
8 0
The answer to your question is 469.7 g
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A LA TEMPERATURA DE 35°C, UNA MUESTRA DE DIOXIDO DE CARBONO OCUPA UN VOLUMEN DE 350 ML. ¿Qué CAMBIO DE VOLUMEN SE PRODUCIRA SI L
Kisachek [45]

Answer:

New volume = 150 mL

Explanation:

Initial temperature, T₁ = 35°C

Initial volume, V₁ = 350 mL

We need to find the change in volume when the temperature drops to 15°C.

The relation between the temperature and the volume is given by Charle's law. Let new volume is V₂. It can be given by :

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{350\times 15}{35}\\\\V_2=150\ mL

So, the new volume is 150 mL.

8 0
2 years ago
List the bonding pairs H and I; S and O; K and Br; Si and Cl, H and F; Se and S; C and H in order of increasing covalent charact
stellarik [79]

Answer:

1. (S,O) < (Se,S) < (C,H) = (H,I) = (H,F) < (Si,Cl) < (K,Br)

Explanation:

The covalent character always increases down the group, this is because ionic character decreases down the group and also electronegativity.

In the same way, Covalent character always decreases across a period because electronegativity increases across a period.

The higher the electronegativity values between the two atoms, the more ionic it will be.

5 0
3 years ago
Why do atoms exchange or share electrons during bonding
kondaur [170]

Answer:

Because it gives them a full valence shell.

Explanation:

5 0
2 years ago
Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp
Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}

For calculating edge length,

a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}

For calculating M_{ave}, we use the formula

M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}

Similarly for calculating (\rho)_{ave}, we use the formula

\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}

From the periodic table, masses of the two elements can be written

M_{Fe}= 55.85g/mol

M_{V}=50.941g/mol

Specific density of both the elements are

(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3

Putting  M_{ave} and \rho_{ave} formula's in edge length formula, we get

a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}}  \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}}  \right )}  \right ]^{1/3}

a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}}  \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}}  \right )}  \right ]^{1/3}

By calculating, we get

a=2.89\times10^{-8}cm=0.289nm

7 0
2 years ago
20.352 mL of chlorine under a pressure of 680. mm Hg are
Lunna [17]

Answer:

0.01144L or 1.144x10^-2L

Explanation:

Data obtained from the question include:

V1 (initial volume) = 20.352 mL

P1 (initial pressure) = 680mmHg

P2 (final pressure) = 1210mmHg

V2 (final volume) =.?

Using the Boyle's law equation P1V1 = P2V2, the volume of the container can be obtained as follow:

P1V1 = P2V2

680 x 20.352 = 1210 x V2

Divide both side by 1210

V2 = (680 x 20.352)/1210

V2 = 11.44mL

Now we need to convert 11.44mL to L in order to obtain the desired result. This is illustrated below:

1000mL = 1 L

11.44mL = 11.44/1000 = 0.01144L

Therefore the volume of the container is 0.01144L or 1.144x10^-2L

7 0
2 years ago
Read 2 more answers
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