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2 years ago

Mixing baking soda and vinegar is an example of an endothermic reaction. This means...

Chemistry

2 answers:

andrew11 [14]2 years ago

7 0

Answer:

it is either:

C or A

Explanation:

i think: C

Murrr4er [49]2 years ago

6 0

Answer: think a

Explanation:

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2. In the water cycle, lake water will do which

aalyn [17]

Water is always on the move. Rain falling today may have been water in a distant ocean days before. And the water you see in a river or stream may have been snow on a high mountaintop. Water is in the atmosphere, on the land, in the ocean, and underground. It moves from place to place through the water cycle.

Where's the water?

There are about 1.4 billion km3 of water (336 million mi3 of water) on Earth. That includes liquid water in the ocean, lakes, and rivers. It includes frozen water in snow, ice, and glaciers, and water that’s underground in soils and rocks. It includes the water that’s in the atmosphere as clouds and vapor.

If you could put all that water together – like a gigantic water drop – it would be 1,500 kilometers (930 miles) across.

6 0

3 years ago

What atomic or hybrid orbitals make up the sigma bond between Cl and F in chlorine trifluoride, ClF3

Umnica [9.8K]

Answer:

sp³d¹ hybridization

Explanation:

Given Cl as central element with three F substrates ...

The VSEPR structure indicates 5 hybrid orbitals that contain 2 diamagnetic orbitals (non-bonded e⁻-pairs) and 3 paramagnetic orbitals (single, non-paired electron for covalent bonding with fluorine) giving a trigonal bypyrimidal parent with a T-shaped geometry.

Valence bond theory predicts the following during bonding:

Cl:[Ne]3s²3p²p²p¹3d⁰

=> [Ne]3s²p²p¹p¹d¹

=> [Ne]3(sp³d)²(sp³d)²(sp³d)¹(sp³d)¹(sp³d)¹

giving 3 ( [Cl](sp³d) - [F]2p¹ ) sigma bonds and 2 non-bonded pairs on Cl.

Note the following images:

Non-bonded electron pairs are in plane of parent geometry and Fluorides covalently bonded to central element Chloride forming the T-shaped geometry.

7 0

3 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago

Your sick puppy is diagnosed with roundworms. They grow inside of the puppy. What type of symbiotic relationship do the puppy an

Ratling [72]

D. Parasitism , the roundworm benefits from the relationship, but the puppy is harmed

4 0

3 years ago

How many molecules are in 2.00 moles of h2o

Sedaia [141]

1 mole = 6.022×10^23 atoms. 1 water molecule = 2 Hydrogen atoms + 1 oxygen atom. So, 1 mole H2O = 1.2044×10^24 hydrogen atoms. Therefore 2 mole H2O will have 2.4088×10^24 hydrogen atoms.

8 0

3 years ago