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mote1985 [20]
2 years ago
8

Such high amounts of pressure cause this layer to remain in a __________ state of matter even though the nickel and iron are at

such a high temperature.
Chemistry
1 answer:
Mkey [24]2 years ago
4 0
Such high amounts of pressure cause this layer to remain in a _____solid_____ state of matter even though the nickel and iron are at such a high temperature.
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You want to determine the effect of a certain fertilizer on the growth of orchids grown in a greenhouse.
Aleks [24]
Hypothesis - I predict that the orchids will grow best with a medium concentration of fertilizer
3 0
2 years ago
What mass of HBO2 is produced from the<br> combustion of 139.5 g of B2H6?<br> Answer in units of g.
aliina [53]

Answer:

m_{HBO_2}=441.8gHBO_2

Explanation:

Hello there!

In this case, since the combustion of B2H6 is:

B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O

Thus, since there is 1:2 mole ratio between the reactant and product, the produced grams of the latter is:

m_{HBO_2}=139.5gB_2H_6*\frac{1molB_2H_6}{27.67gB_2H_6} *\frac{2molHBO_2}{1molB_2H_6} *\frac{43.82gHBO_2}{1molHBO_2}

m_{HBO_2}=441.8gHBO_2

Best regards!

5 0
2 years ago
In a first-order decomposition reaction, 50.0% of a compound decomposes in 10.5 min.
mihalych1998 [28]

In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol t_{1/2}.

(a) For first order reaction, rate constant and half life time are related to each other as follows:

k=\frac{0.6932}{t_{1/2}}=\frac{0.6932}{10.5 min}=0.066 min^{-1}

Thus, rate constant of the reaction is 0.066 min^{-1}.

(b) Rate equation for first order reaction is as follows:

k=\frac{2.303}{t_{1/2}}log\frac{[A_{0}]}{[A_{t}]}

now, 75% of the compound is decomposed, if initial concentration [A_{0} ] is 100 then concentration at time t [A_{t} ] will be 100-75=25.

Putting the values,

0.066 min^{-1}=\frac{2.303}{t}log\frac{100}{25}=\frac{2.303}{t}(0.6020)

On rearranging,

t=\frac{2.303\times 0.6020}{0.066 min^{-1}}=21 min

Thus, time required for 75% decomposition is 21 min.

8 0
3 years ago
Problem 12.002 the molar analysis of a gas mixture at 30°c, 2 bar is 40% n2, 50% co2, 10% ch4. determine
andreev551 [17]
The problem is incomplete. However, there can only be two probable questions for this problem. First, you can be asked the individual partial pressures of each gas. Second, you can be asked the volume occupied by each gas. I can answer both cases for you.

1.

Let's assume ideal gas.
Pressure for N₂: 2 bar*0.4 = 0.8 bar
Pressure for CO₂: 2 bar*0.5 = 1 bar
Pressure for CH₄: 2 bar*0.1 = 0.2 bar

2. For the volume, let's find the total volume first.

V = nRT/P = (1 mol)(8.314 J/mol-K)(30 +273 K)/(2 bar*10⁵ Pa/1 bar)
V = 0.0126 m³
Hence,
Volume for N₂: 0.0126 bar*0.4 = 0.00504 m³
Volume for CO₂: 0.0126*0.5 = 0.0063 m³
Volume for CH₄: 0.0126*0.1 = 0.00126 m³
3 0
2 years ago
El oxígeno y el azufre reaccionan con el cobre para formar óxido de cobre (CuO) y sulfuro de cobre (CuS), respectivamente. ¿Qué
oksian1 [2.3K]
Yes it need to be like that cause when it like that it like that
5 0
2 years ago
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