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3 years ago

Pleas help me i have to submit homework i hour i will mark brainliest

Chemistry

1 answer:

steposvetlana [31]3 years ago

6 0

Answer:

I can help

Explanation:

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What range on the glycemic index (GI) corresponds to a high glycemic impact food? 70 and above 56 to 69 1 to 55 1 to 35

german

Answer: 70 and above

Explanation:

This is a fact.

8 0

1 year ago

MgNH4PO4•6H2O loses H2O in a stepwise fashion as it is heated. Between 40 degrees celsius and 60 degrees celsius, the monohydrat

tino4ka555 [31]

Answer:

$\%\ Composition\ of\ phosphorus=20.00\ \%$

Explanation:

The molecular formula of the monohydrate formed = $MgNH_4PO_4.H_2O$

The molecular mass of the monohydrate formed = $Mass_{Mg}+Mass_{N}+6\times Mass_{H}+5\times Mass_O$

So, Mass = 24 + 14 + 6 × 1 + 5 × 16 = 155 g

Mass of phosphorus = 31 g

Thus,

$\%\ Composition\ of\ phosphorus =\frac{Mass_{phosphorus}}{Total\ mass}\times 100$

$\%\ Composition\ of\ phosphorus=\frac{31}{155}\times 100$

$\%\ Composition\ of\ phosphorus=20.00\ \%$

4 0

3 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

3 0

3 years ago

The reaction between an organic acid and an alcohol produces

NARA [144]

If it’s a carboxylic acid with an alcohol then it’ll form an ester and water.

5 0

3 years ago

Please Please Please help + show the work please! :)

Mazyrski [523]

The gram-formula mass of Sm is 150.36 u, and the gram-formula mass of O is 15.999 u, so the gram-formula mass of SmO is about 150.36+15.999 = 160.36 g/mol.

So, there are about (9.30 * 10^-3)(160.36)=1.49 grams

4 0

2 years ago

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