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3 years ago

Lol I don’t know (,:

Chemistry

2 answers:

Tju [1.3M]3 years ago

8 0

Answer:

Wash them with water and soap

Explanation:

The other just put like a shield around your hands but water and soap kills bacteria

Romashka-Z-Leto [24]3 years ago

5 0

Answer:

the best way to wash your hands is to WASH THEM WITH SOAP AND WATER

Explanation:

BECAUSE soap and water will keep your hands clean clean for a long time

You might be interested in

45.5 g of ice is heated from -25 degrees celsius to 150 degrees celcius. Calculate the total amount of heat required for the ent

tino4ka555 [31]

Answer:

Q = 16163.88 Joules

Explanation:

Given the following data;

Initial temperature, T1 = -25°C

Final temperature, T2 = 150°C

Mass = 45.5 g

Specific heat capacity of ice = 2.03 J/g°C

To find the quantity of heat required;

Heat capacity is given by the formula;

$Q = mcdt$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 150 - (-25)

dt = 150 + 25

dt = 175°C

Substituting into the formula, we have;

$Q = 45.5 * 2.03 * 175$

Q = 16163.88 Joules

4 0

2 years ago

Simple question ill name u brainiest

BigorU [14]

Answer:

Test tube :)

Explanation:

4 0

3 years ago

Read 2 more answers

Melissa's mother was making chocolate chip cookies, but told Melissa that she shouldn't eat the raw cookie dough. Why is it unsa

sashaice [31]

the answer to the question is D

4 0

3 years ago

calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o

Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of $Na_3PO_4$ = 12.00 g

Mass of $CaCl_2$ = 10.0 g

Molar mass of $Na_3PO_4$ = 164 g/mol

Molar mass of $CaCl_2$ = 111 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

Molar mass of $Ca_3(PO_4)_2$ = 310 g/mol

First we have to calculate the moles of $Na_3PO_4$ and $CaCl_2$ .

$\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}$

$\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol$

and,

$\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}$

$\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

From the balanced reaction we conclude that

As, 3 mole of $CaCl_2$ react with 2 mole of $Na_3PO_4$

So, 0.0901 moles of $CaCl_2$ react with $\frac{2}{3}\times 0.0901=0.0601$ moles of $Na_3PO_4$

From this we conclude that, $Na_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$ and $Ca_3(PO_4)_2$

From the reaction, we conclude that

As, 3 mole of $CaCl_2$ react to give 6 mole of $NaCl$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{6}{3}\times 0.0901=0.1802$ mole of $NaCl$

and,

As, 3 mole of $CaCl_2$ react to give 1 mole of $Ca_3(PO_4)_2$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{1}{3}\times 0.0901=0.030$ mole of $Ca_3(PO_4)_2$

Now we have to calculate the mass of $NaCl$ and $Ca_3(PO_4)_2$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

$\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g$

and,

$\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2$

$\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g$

Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

5 0

3 years ago

the pressure and temperature of 10 liters of gas are doubled. if the original condition are 2 atmosphere of pressure and 400k wh

Harman [31]

Answer:

$\boxed{\text{10 L}}$

Explanation:

We have two pressures, two temperatures, and one volume.

This looks like a question in which we can use the Combined Gas Law to calculate the volume.

$\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}$

Data:

$\begin{array}{rclrclrcl}p_{1}& =& \text{2 atm}\qquad & V_{1} &= & \text{10 L}\qquad & T_{2}& =& \text{400 K}\\p_{2}& =& \text{4 atm}\qquad & V_{2} &= & \text{?}\qquad & T_{2}& =& \text{800 K}\\\end{array}$

Calculation:

$\begin{array}{rcl}\dfrac{2 \times 10}{400}& =& \dfrac{4V_{2} }{800}\\\\0.050& = &0.0050V_{2}\\V_{2}& = &\mathbf{10 L}\end{array}\\\text{The final volume is }\boxed{\textbf{10 L}}$

4 0

3 years ago