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Nataliya [291]
3 years ago
14

The rising magma that may result from this type of plate movement may produce HELP

Chemistry
1 answer:
nirvana33 [79]3 years ago
7 0

Volcanic islands, seamounts, guyouts

Explanation:

The magma rising from this type of plate movement may produce under water volcanic activities.

The magma here would be mafic - ultramafic because it is produced from melting of oceanic plates.

As the subducting oceanic plate melts and begins to rise, it will settle and cool on the ocean floor resulting in the formation of massive and extensive seamounts.

With time, they outgrow the ocean level and forms volcanic Islands.

learn more:

lithosphere brainly.com/question/9582362

#learnwithBrainly

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LuckyWell [14K]
It is 0 gradus in celcius
6 0
2 years ago
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WILL GIVE 50 POINTS FOR ALL ANSWERS!!! PLEASE HELP!
blondinia [14]

Answer:

the answer would be (A.) and (D.).

Explanation:

the reason for that being is because if calcium sulfate is a main component of plaster of  paris you would need to find out what is in it that makes it the main component aka (the formula) therefore part of the answer is (A.). The other part of the answer was (D.) because you would need to find the amount of calcium sulfate that contain 12 grams of oxegeon atoms because you finding the answer to that could lead to the answer of what is the main component of plaster of Paris.

3 0
2 years ago
Estimate your de Broglie wavelength when you are running. (For this problem use h = 10^−34 in SI units and 1 lb is equivalent to
AlekseyPX

Answer:

A. your running speed 1.5 m/s

B. your mass 70 kg

C. your de Broglie wavelength 6.32x10^{-36}m

Explanation:

Hello there!

In this case, since the equation for the calculation of the Broglie wavelength is:

\lambda =\frac{h}{m*v}

We can assume a running speed of about 1.5 m/s and a mass of 70 kg, so the resulting Broglie wavelength is:

\lambda =\frac{6.626x10^{-34}kg\frac{m}{s} }{70kg*1.5\frac{m}{s} }\\\\\lambda =6.32x10^-36m

Best regards!

7 0
2 years ago
Copper oxide, CuO, reacts with hydrochloric acid, HCI, to produce copper chloride, CuCL2 and water
spayn [35]

Explanation:

El óxido de cobre (II), también llamado antiguamente óxido cúprico ({\displaystyle {\ce {CuO}}}{\displaystyle {\ce {CuO}}}), es el óxido de cobre con mayor número de oxidación. Como mineral se conoce como tenorita.

{\displaystyle {\ce {2Cu + O2 = 2CuO}}}{\displaystyle {\ce {2Cu + O2 = 2CuO}}}

Aquí, se forma junto con algo de óxido de cobre (I) como un producto lateral, por lo que es mejor prepararlo por calentamiento de nitrato de cobre (II), hidróxido de cobre (II) o carbonato de cobre (II):

{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}

{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}

{\displaystyle {\ce {CuCO3 = CuO + CO2}}}{\displaystyle {\ce {CuCO3 = CuO + CO2}}}

El óxido de cobre (II) es un óxido básico, así se disuelve en ácidos minerales tales como el ácido clorhídrico, el ácido sulfúrico o el ácido nítrico para dar las correspondientes sales de cobre (II):

{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}

{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}

{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}

Reacciona con álcali concentrado para formar las correspondientes sales cuprato.

{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}

Puede reducirse a cobre metálico usando hidrógeno o monóxido de carbono:

{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}

{\displaystyle {\ce {CuO + CO = Cu + CO2}}}{\displaystyle {\ce {CuO + CO = Cu + CO2}}}

6 0
2 years ago
PLEASE HELP!!!!!!
pogonyaev

Answer:

m_{B}^{theoretical}=0.365gB

Y=87.1\%

Explanation:

Hello there!

In this case, since the reaction (A->B) have an initial amount of pure 4-aminobenzoic acid, the first step to compute the theoretical yield is to solve the following stoichiometric setup:

m_{B}^{theoretical}=0.303gA*\frac{1molA}{137.14gA}*\frac{1molB}{1molA}*\frac{165.19 gB}{1molB}\\\\   m_{B}^{theoretical}=0.365gB

Whereas A stands for 4-aminobenzoic acid and B for the benzocaine. Moreover, we compute the percent yield by dividing the actual yield (0.318 g) by the theoretical one (0.365 g):

Y=\frac{0.318g}{0.365g} *100\%\\\\Y=87.1\%

Best regards!

4 0
2 years ago
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