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3 years ago

Plants use photosynthesis,sunlight,water and carbon dioxide to prepare their food. What are they called in a group?

1 answer:

3 0

Answer: Plants are autotrophs, which means they produce their own food. They use the process of photosynthesis to transform water, sunlight, and carbon dioxide into oxygen, and simple sugars that the plant uses as fuel.

Explanation:

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A potassium chloride solution that also contains 5% (m/V) dextrose is administered intravenously to treat some forms of malnutri

alexgriva [62]

ci=oncentration of pottasium = 3.5 * 10^-2 mol/litre

3 0

3 years ago

Read 2 more answers

Write balanced net ionic equations and the corresponding equilibrium equations for the stepwise dissociation of the triprotic ac

Elza [17]

Answer:

$H_3PO_4(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+H_2PO_4^{-}(aq); \ Ka_1$

$H_2PO_4^{-}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+HPO_4^{-2}(aq); \ Ka_2$

$HPO_4^{-2}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+PO_4^{-3}(aq); \ Ka_3$

Explanation:

Hello!

In this case, since the phosphoric acid is a triprotic acid, we infer it has three stepwise ionization reactions in which one hydrogen ion is released at each step, considering they are undergone due to the presence of water, thus, we proceed as follows:

$H_3PO_4(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+H_2PO_4^{-}(aq); \ Ka_1$

$H_2PO_4^{-}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+HPO_4^{-2}(aq); \ Ka_2$

$HPO_4^{-2}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+PO_4^{-3}(aq); \ Ka_3$

Moreover, notice each step has a different acid dissociation constant, which are quantified in the following order:

Ka1 > Ka2 > Ka3

Best regards!

3 0

3 years ago

Why is the melting of ice a physical change? It changes the chemical composition of water. It does not change the chemical compo

snow_lady [41]

It does not change the chemical in the composition of water. hope it helps

3 0

3 years ago

Read 2 more answers

Which of the following has parts a and b labeled correctly

Alex

Answer:

the last one

Explanation:

A is the solution and the solvent and B is the solute

4 0

3 years ago

If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?

ElenaW [278]

Answer:

a. 1810mL

Explanation:

When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$ where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation: $T_C+273=T_K$

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, $T_1 = 273[K]$ , and $T_2 = (49.4+273)[K]=322.4[K]$ .

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

$\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}$

$\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})$

$1810.04571428[mL]=V_2$

Adjusting for significant figures, this gives $V_2=1810[mL]$

4 0

2 years ago