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3 years ago

HELP IM SCARED!!! will a wasp die to the smell of bleach? THERE'S A WASP IN MY BATHROOM!!!!

Chemistry

2 answers:

lesantik [10]3 years ago

6 0

Answer:

Explanation:

It can die but not quickly. You can get a spray for wasp it works great I try it and kills them. But don't put bleach on them or makes bleach vapor may be harmful for you.

Nesterboy [21]3 years ago

4 0

no, u will have to soak then in order for them to die

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How to do Criss cross method of lithium carbide?

timama [110]

You will need to go on yt

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2 years ago

By mistake, you added salt instead of sugar to the<br> oil. How can you remove the salt?

Jet001 [13]

Answer:

by adding water into the mix

Explanation:

this will dissolve the salt

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2 years ago

Read 2 more answers

What is the pH of a KOH solution that has [H ] = 1. 87 × 10–13 M? What is the pOH of a KOH solution that has [OH− ] = 5. 81 × 10

vlada-n [284]

pH is the hydrogen ion concentration and pOH is the hydroxide ion concentration in the solution. pH KOH is 12.73, pOH KOH is 2.24 and pH NaCl is 7.

pH is the negative log of the hydrogen ion concentration and pOH is the negative log of the hydroxide ion concentration.

The relation between the pH and pOH can be given as, $\rm pOH = 14 - pH$

The pH of KOH can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the first case, the concentration of the KOH is $1. 87 \times 10^{-13}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 87 \times 10^{-13}\;\rm M ]\\\\&= 12.73\end{aligned}$

Hence, the pH of KOH is 12.73.

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pOH of KOH can be calculated by the formula,

$\rm pOH = \rm -log [OH^{-}]$

The hydroxide concentration of the KOH solution is $5. 81 \times 10^{-3}\;\rm M$

Substituting value in the equation:

$\begin{aligned} \rm pOH &= \rm -log [OH^{-}]\\\\&= \rm -log [5. 81 \times 10^{-3}\;\rm M ]\\\\&= 2.24 \end{aligned}$

Hence, the pOH of KOH is 2.24

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The pH of NaCl can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the third case, the concentration of the NaCl is $1. 00\times 10^{-7}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 00 \times 10^{-7}\;\rm M ]\\\\&= 7 \end{aligned}$

Hence, the pH of KOH is 7.0.

Therefore, KOH is basic and NaCl is approximately neutral.

Learn more about pH and pOH here:

brainly.com/question/13885794

3 0

2 years ago

Think about the lab procedure you just read. Label each factor below with V for “variable” or C for “constant”.

luda_lava [24]

<h2>Answer:</h2><h3>The temperature of the gas: V</h3>

The temperature of gas is a variable quantity. It can be changed by changing energy or pressure of gas.

<h3>The amount of gas in the tube (in terms of mass and moles): C</h3>

It is a constant entity. As mass of gas once taken can not be changed by changing temperature, pressure etc.

<h3>The radius of the tube: C</h3>

The radius of tube cannot change at any rate.

<h3>The temperature of the gas (changed by the water surrounding it): V</h3>

It can be changed by changing the temperature of water surrounding it.

It can never be changed.

<h3>The pressure of the gas: V</h3>

It can be changed by simply changing temperature and volume of gas.

8 0

3 years ago

Read 2 more answers

Determine the boiling point of a 3.70 m solution of phenol in benzene. Benzene has a boiling point of 80.1°C and a boiling point

xeze [42]

Answer: The boiling point of a 3.70 m solution of phenol in benzene is $89.5^0C$

Explanation:

Elevation in boiling point:

$\Delta T_b=i\times k_b\times m$

where,

$\Delta T_b$ = change in boiling point

i= vant hoff factor = 1 (for benzene which is a non electrolyte )

$k_b$ = boiling point constant = $2.53^0C/kgmol$

m = molality = 3.70

$T_{solution}-T_{solvent}=i\times k_b\times m$

$T_{solution}-80.1^0C=1\times 2.53\times 3.70$

$T_{solution}=89.5^0C$

Thus the boiling point of a 3.70 m solution of phenol in benzene is $89.5^0C$

8 0

2 years ago