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2 years ago

Can someone answer these questions about the LACK OF CLEAN DRINKING WATER? PLS. I will give 30 pts.

Chemistry

1 answer:

Andru [333]2 years ago

8 0

Answer:

Answer

Explanation:

Zack Gomez

Lack of clean water

Water should be clean cause we need water so we can drink it cause we live off of it so we need it so it should be clean. Animals need it to survive in the hot so they need it so it should be clean. Insects need it to survive so they need it so it should be clean. THE END

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A chemist wants to observe the following reaction:

stiv31 [10]

It take more energy to break the bonds of the reactants and less energy is given off when the product bonds are formed.

<h3>What is Energy?</h3>

Energy is defined as the ability to do work. Work is done in the breaking or formation of bonds.

The standard Enthalpy (ΔH) of water which was formed in the given reaction is negative.

ΔH= Δproduct - Δreactant

This means that the energy to break the bonds of the reactants is more.

Read more about Enthalpy here brainly.com/question/14291557

3 0

2 years ago

Is carbon dioxide a compound, element,or a mixture? wat about air wat is it

sveta [45]

Carbon dioxide is a compound. Oxygen is an element. A compound is 2 or more elements chemically combined. An element is a pure substance. A mixture is combining two or more different compounds/elements.

6 0

3 years ago

How many grams of a 14.0% (w/w) sugar solution contain 62.5<br> of sugar?

Delicious77 [7]

Answer:

$m_{solution}=446.4g$

Explanation:

Hello,

In this case, by-mass percent is computed in terms of the mass of the solute and the mass of the solvent as shown below:

$\% m=\frac{m_{solute}}{m_{solution}} *100\%$

Thus, solving for the mass of the solution, we obtain:

$\frac{x}{y} m_{solution}=\frac{m_{solute}}{\% m}*100\%\\ \\m_{solution}=\frac{62.5g}{14.0\%}*100\% \\\\m_{solution}=446.4g$

Regards.

3 0

3 years ago

Consider 4.00 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.80 L and the temperature is increased to 3

stealth61 [152]

This is an exercise in<u> the General Combined Gas Law</u>.

To start solving this exercise, we obtain the following data:

V₁ = 4.00 l
P₁ = 365 mmHg
T₁ = 20 °C + 273 = 293 K
V₂ = 2,80 l
T₂ = 30 °C + 273 = 303 K
P₂ = ¿?

We apply the following formula:

P₁V₁T₂=P₂V₂T₁ ⇒ General formula

Where:

P₁=Initial pressure
V₁=Initial volume
T₂=end temperature
P₂=end pressure
T₂=end temperature
V₁=Initial temperature

We clear for final pressure (P2)

$\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{P_{1}V_{1}T_{2}}{V_{2}T_{1}} \ \ \to \ \ \ Formula \end{gathered}$}$

We substitute our data into the formula:

$\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{(365 \ mmHg)(4.00 \not{l})(303 \not{K})}{(2.80 \not{l})(293\not{K})} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{442380 \ mmHg}{ 820.4 } \end{gathered}$}$

$\boxed{\large\displaystyle\text{$\begin{gathered}\sf P_{2}=539.224 \ mmHg \end{gathered}$}}$

Answer: The new canister pressure is 539.224 mmHg.

<h2>{ Pisces04 }</h2>

6 0

2 years ago

Sr 32.8 g, Si 5.2 g and O 11.9 g empirical formula number 11

horsena [70]

$\boxed{\begin{array}{c|c|c|c|c}\boxed{\sf Element}&\boxed{\sf Mass\:in\: compound}&\boxed{\sf No\:of\:moles}&\boxed{\sf Ratio}&\boxed{\sf Simplified\:ratio}\\ \sf Sr &\sf 32.8g &\sf \dfrac{32.8}{87}=0.37&\sf \dfrac{0.37}{0.18}=2.05&\sf 2 \\ \sf Si &\sf 5.2g &\sf \dfrac{5.2}{28}=0.18 &\sf \dfrac{0.18}{0.18}=1&\sf 1\\ \sf O&\sf 11.9g &\sf \dfrac{11.9}{16}=0.74&\sf\dfrac{0.74}{0.18}=4.1 &\sf 4\end{array}}$

Empirical formula:-

$\\ \sf\longmapsto SiSr_2O_4$

8 0

2 years ago