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2 years ago

What's the agriculture

Chemistry

2 answers:

trapecia [35]2 years ago

4 0

Answer:

the science or practice of farming, including cultivation of the soil for the growing of crops and the rearing of animals to provide food, wool, and other products.

Explanation:

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ivann1987 [24]2 years ago

3 0

Answer:

Agriculture is a practice of farming, including cultivation of the soil for growing crops and the rearing of animals to provide food,wool and others.

<h2>Hope it helps !</h2>

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Need chemistry help look on the picture answer all questions please

pychu [463]

1. Aqeuous or dissolved
2.AgNO3+ NaCl are reactants
3. NaNO3 is precipitate
4. Reaction is endothermic(Delta stands for Heat)
5. Neither a reactant nor a product, MnO2 is a catalyst

Please Mark Brainliest. Hope the Answer helps

6 0

2 years ago

Which of these statements is correct?

Crazy boy [7]

Answer:

Explanation:

3 0

2 years ago

Calculate the EMF between copper and silver Ag+e-E=0.89v<br>Cu=E=0.34v

bogdanovich [222]

Answer:

Depending on the $E^\circ$ value of $\rm Ag^{+} + e^{-} \to Ag\; (s)$ , the cell potential would be:

$0.55\; \rm V$ , using data from this particular question; or
approximately $0.46\; \rm V$ , using data from the CRC handbooks.

Explanation:

In this galvanic cell, the following two reactions are going on:

The conversion between $\rm Ag\; (s)$ and $\rm Ag^{+}$ ions, $\rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s)$ , and
The conversion between $\rm Cu\; (s)$ and $\rm Cu^{2+}$ ions, $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ .

Note that the standard reduction potential of $\rm Ag^{+}$ ions to $\rm Ag\; (s)$ is higher than that of $\rm Cu^{2+}$ ions to $\rm Cu\; (s)$ . Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if $\rm Ag^{+}$ ions are reduced while $\rm Cu\; (s)$ is oxidized.

Therefore:

The reduction reaction at the cathode will be: $\rm Ag^{+} + e^{-} \to Ag\; (s)$ . The standard cell potential of this reaction (according to this question) is $E(\text{cathode}) = 0.89\; \rm V$ . According to the 2012 CRC handbook, that value will be approximately $0.79\; \rm V$ .

The oxidation at the anode will be: $\rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}$ . According to this question, this reaction in the opposite direction ( $\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)$ ) has an electrode potential of $0.34\; \rm V$ . When that reaction is inverted, the electrode potential will also be inverted. Therefore, $E(\text{anode}) = -0.34\; \rm V$ .

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}$ .

Using data from the 1985 and 2012 CRC Handbook:

$\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}$ .

5 0

3 years ago

The combustion of glucose (c6h12o6) with oxygen gas produces carbon dioxide and water. this process releases 2803 kj per mole of

V125BC [204]

Answer:- 335 kcal of heat energy is produced.

Solution:- The balanced equation for the combustion of glucose in presence of oxygen to give carbon dioxide and water is:

$C_6H_1_2O_6+6O_2\rightarrow 6CO_2+6H_2O$

From given info, 2803 kJ of heat is released bu the combustion of 1 mol of glucose. We need to calculate the energy produced when 3.00 moles of oxygen react with excess of glucose.

We could solve this using dimensional analysis as:

$3.00mol O_2(\frac{1mol glucose}{6mol O_2})(\frac{2803 kJ}{1mol glucose})$

= 1401.5 kJ

Now, let's convert kJ to kcal.

We know that, 1kcal = 4.184kJ

So, $1401.5kJ(\frac{1kcal}{4.184kJ})$

= 335 kcal

Hence, 335 kcal of heat energy is produced by the use of 3.00 moles of oxygen gas.

8 0

2 years ago

What would be the mass in grams of 1.204 x 1024 molecules of sulfur dioxide

melomori [17]

Answer:

mass of sulfur = 96 g

Explanation:

no of moles of sulfur dioxide in $1.204\times 10^{24}$ molecules = $\frac{1.204\times 10^{24}}{avagadro number }= \frac{1.204\times 10^{24}}{6.023\times 10^{23}}$

= 2 moles

therefore mass of sulfur dioxide = moles×atomic number

=2×(16+32)

=96

6 0

3 years ago

What's the agriculture​

What's the agriculture