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3 years ago

Help me, I’m bad a science.

Chemistry

1 answer:

Sauron [17]3 years ago

8 0

Answer:

Removing the worms is they first key to overcoming this problem. Thus you should ensure that the worms have not spread throughout the house/facility, or caused more damage.

Secondly, you should replace the pipes taht the worms got into, or clean them at most. If it’s simply worm droppings then just clean it, but if it’s more and worse than that alike slime then it is best to remove and replace it.

the only way the worms escaping could affect something is if they decontaminated it, or alas ate or destroyed it. Other than that if they died you could simply replace them, but e things cannot be replaced

Explanation:

You didn’t actually tell us any details so that the best I got from that

can I have brainlest pwease?

You might be interested in

Did I get This right I’m confused

Paul [167]

Hey bud !
You were close congrats !
the only problem i saw was the units part at the end.
remember volume is 3 dimensional
length width and height
so it is m^3

3 0

4 years ago

How to determine reaction order in reactant?

Artyom0805 [142]

In order to determine, Order of reaction, we have to add all the exponents written in the Chemical form, on the Reactant species.

Hope this helps!

8 0

3 years ago

A compound was found to be soluble in water. It was also found that addition of acid to an aqueous solution of this compound res

gogolik [260]

Answer:

c. Cr

Explanation:

$Cr^3+(aq) + CO_3^2-(aq) ------------> Cr_2CO_3(s)$

The compound is containing $CO_3^2-$ ion.

If acid reacts with it, CO_2 evolves.

$H^+ + CO_3^2- ----------> H_2O + CO_2$

therefore, Cr would form a precipitate when added to an aqueous solution of this compound. Cr2CO3 is the precipitate.

3 0

3 years ago

The analysis of a hydrocarbon revealed that it was 85.7% C and 14.3% H by mass. When 1.77 g of the gas was stored in a 1.500-L f

gtnhenbr [62]

Answer:

The formula of hydrocarbon = $C_3H_6$

Explanation:

$Moles =\frac {Given\ mass}{Molar\ mass}$

% of C = 85.7

Molar mass of C = 12.0107 g/mol

% moles of C = 85.7 / 12.0107 = 7.14

% of H = 14.3

Molar mass of H = 1.00784 g/mol

% moles of H = 14.3 / 1.00784 = 14.19

Taking the simplest ratio for C and H as:

7.14 : 14.19 = 1 : 2

The empirical formula is = $CH_2$

Also, Given that:

Pressure = 508 Torr

Temperature = 17 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (17 + 273.15) K = 290.15 K

Volume = 1.500 L

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 62.3637 L.torr/K.mol

Applying the equation as:

508 Torr × 1.500 L = n × 62.3637 L.torr/K.mol × 290.15 K

⇒n = 0.0421 moles

Given that :

Amount = 1.77 g

Molar mass = ?

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0421\ moles= \frac{1.77\ g}{Molar\ mass}$

Molar mass of the hydrocarbon = 42.04 g/mol

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12 + 2×1= 14 g/mol

Molar mass = 42.04 g/mol

So,

Molecular mass = n × Empirical mass

42.04 = n × 14

⇒ n = 3

<u>The formula of hydrocarbon = $C_3H_6$ </u>

5 0

3 years ago

A hypothetical covalent molecule, X–Y, has a dipole moment of 1.93 1.93 D and a bond length of 109 pm. 109 pm. Calculate the par

Gnesinka [82]

Answer:

q= 110.5 ke

Explanation:

Dipole moment is the product of the separation of the ends of a dipole and the magnitude of the charges.

μ = q * d

μ= Dipole moment (1.93 D)

q= partial charge on each pole

d= separation between the poles(109 pm).

e= electronic charge ( 1.60217662 × 10⁻¹⁹ coulombs)

So,

q= $\frac{1.93}{109 * 10^{-12} }$ coulombs

q = $\frac{1.93}{109 * 10^{-12} * 1.60217662 * 10^{-19} }$ e

q = 1.105 * 10⁵ e

q= 110.5 ke

4 0

3 years ago