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3 years ago

If I have 251.14 grams of citric acid and calcium bicarbonate mixture how much citrus acid do I have??

1 answer:

6 0

About 255.3 grams is the answer I believe. I just had a school work packet and had that question on it, and that is the answer that I had put on it I believe.

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What are chemical changes of sodium hydroxide

pochemuha

Answer: Sodium + water → hydrogen gas + sodium hydroxide (aq) + heat [(aq) means “dissolved in water”. It stands for “aqueous”.] If we boil off the water, we will be able to see the sodium hydroxide crystals. The sodium hydroxide is one of the two new substances produced by this chemical change.

8 0

3 years ago

What adaptations allow a camel and a cactus to survive in warm environments?

lesya [120]

Answer:

A camel stores fat in its hump, while the cactus stores water in its thick stem.

Explanation:

8 0

3 years ago

Read 2 more answers

CdF2(s)⇄Cd2+(aq)+2F−(aq)A saturated aqueous solution of CdF2 is prepared. The equilibrium in the solution is represented above.

kupik [55]

Answer:

The correct answer is option a.

Explanation:

$CdF_2(s)\rightleftharpoons Cd^{2+}(aq)+2F^-(aq)$

Equilibrium concentration cadmium ions = $[Cd^{2+}]=0.0585 M$

Equilibrium concentration fluoride ions = $[F^{-}]=0.117 M$

Molar solubility is the maximum concentration of salt present in water in ionic form beyond that no more salt will exist in its ionic form and will settle down in bottom of the solution.

The molar solubility of the solid cadmium fluoride = 0.0585 M

$CdF_2(s)\rightleftharpoons Cd^{2+}(aq)+2F^-(aq)$ ..[1]

$NaF(s)\rightleftharpoons Na^{+}(aq)+F^-(aq)$

Due to addition of sodium fluoride will increase concentration of fluoride in the solution.And due to common ion effect the equilibrium will shift in backward direction in [1], that is precipitation of more cadmium fluoride.

Hence, decrease in solubility will be observed.

8 0

3 years ago

8. A sample of potassium chlorate (KCIO,) was heated in a test tube and decomposed 2KC?(s) 302 (g) + 2KCIO, (s) The oxygen was c

dangina [55]

Answer:

Partial pressure of $O_{2}$ in the gas was 733 torr and mass of $KClO_{3}$ in the sample was 2.12 g.

Explanation:

a) Total pressure of gas = (partial pressure of water vapour)+(partial pressure of $O_{2}$ )

Here partial pressure of water vapour is 21 torr and total pressure of gas is 754 torr.

So, partial pressure of $O_{2}$ = (total pressure of gas)-(partial pressure of water vapour) = (754 torr) - (21 torr) = 733 torr

b) Lets assume that $O_{2}$ behaves ideally. Hence-

PV=nRT

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin

here P = 733 torr = $(733\times 0.001316)atm$ = 0.9646 atm

V = 0.65 L, R = 0.082 L.atm/(mol.K), T=(273+22)K = 295 K

So, $n=\frac{PV}{RT}$

= $\frac{(0.9646 atm)\times (0.65 L)}{(0.082 L.atm/(mol.K))\times (295 K)}$

= 0.0259 moles

As 3 moles of $O_{2}$ are produced from 2 moles of $KClO_{3}$ therefore 0.0259 moles of $O_{2}$ are produced from $(\frac{2\times 0.0259}{3})$ moles or 0.0173 moles of $KClO_{3}$ .

Molar mass of $KClO_{3}$ = 122.55 g

So mass of $KClO_{3}$ in sample = $(0.0173\times 122.55)g$

= 2.12 g

7 0

3 years ago

Lithium iodide has a lattice energy of -7.3 x 102(kj/mol) and a heat of hydration of -793 kj/mol. find the heat of solution for

marishachu [46]

Given molecule Lithium iodide (LiI)

Heat of hydration = -793 kj/mol

Lattice energy = -730 kJ/mol

Heat of hydration = Heat of solution - Lattice energy

Heat of solution = Hydration + Lattice = -793 + (- 730) = -1523 kJ/mol

Now,

Mass of LiI = 15.0 g

molar mass of LiI = 134 g/mol

# moles of LiI = 15/134 = 0.112 moles

Heat of solution for 1 mole of LiI = -1523 KJ

Therefore, for 0.112 moles of LiI the corresponding heat is

= 0.112 *(-1532) = 171.6 kJ

5 0

3 years ago