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2 years ago

If the reactants of a reaction are HCI and NaOH, what could be a product?

Chemistry

1 answer:

Nutka1998 [239]2 years ago

6 0

Answer

HCl+NaOH=NaCl+H2O (NaCl+H20)=the product

is a double substitution reaction

so the product ,one of them will be or Nacl or Water(H2O)

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Strigol is an important plant hormone that is released by crops such as rice and sugarcane. Unfortunately for these plants, stri

Len [333]

The hydrogen deficiency index( HDI) of strigol is = 10

<h3>How to calculate HDI:</h3>

The hydrogen deficiency index is used to measure the number of degree of unsaturation of an organic compound.

Strigol is an example of an organic compound because it contains carbons and hydrogen.

To calculate the HDI using the molecular formula given (C19H20O6) the formula for HDI is used which is:

$hdi = \frac{1}{2} (2c + 2 + n - h - x)$

where C = number of carbon atoms = 19

n= number of nitrogen atoms = 0

h= number of hydrogen atoms = 20

X = number of halogen atoms = 0

Note that oxygen was not considered because it forms two bonds and has no impact.

There for HDI =

$\frac{1}{2} (2 \times 19 + 2 + 0 - 20 - 0)$

HDI=

$\frac{1}{2} (40 - 20)$

HDI =

$\frac{1}{2} \times 20$

HDI = 10

Therefore, the hydrogen deficiency index of strigol is = 10

Learn more about unsaturated compounds here:

brainly.com/question/490531

7 0

3 years ago

Which of the following represents a chemical change? A.Souring of milk B.Melting of chocolate C.Condensation of water D.Breaking

leonid [27]

Chemical changes occur when a substance combines with another to form a new substance, called chemical synthesis or, alternatively, chemical decomposition into two or more different substances.

So which one do you think is the answer?

8 0

3 years ago

Read 2 more answers

How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

Ksenya-84 [330]

Answer:

a. Approximately $1.3\; \rm mL$ .

b. Approximately $7.2\; \rm mL$ .

Explanation:

The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

However, the volume of both solutions were given in mililiters $\rm mL$ . Convert these volumes to liters:

$\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L$ .

$\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L$ .

In a solution of volume $V$ where the concentration of a solute is $c$ , there would be $c \cdot V$ (moles of) formula units of this solute.

Calculate the number of moles of $\rm NaCl$ formula units in each of the two solutions:

Solution in a.:

$n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol$ .

Solution in b.:

$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

What volume of that $3.4\; \rm M$ (same as $3.4 \; \rm mol \cdot L^{-1}$ ) $\rm NaCl$ solution would contain that many

For the solution in a.:

$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL$ .

Similarly, for the solution in b.:

$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

8 0

3 years ago

All waves carry A) energy. B) light. C) matter. D) particles.

7nadin3 [17]

A. energy would be the answer

4 0

3 years ago

Read 2 more answers

Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low

LenaWriter [7]

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

$\Delta T_f=i\times k_f\times m$ ..[1]

$m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}$

Where:

i = van't Hoff factor

$k_f$ = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: $\Delta T_{f,A}$