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Elden [556K]
3 years ago
15

Select the correct answer. Samuel adds a teaspoon of salt to a glass of water. He notices that the salt disappears. Samuel takes

a sip to discover that the water tastes salty. What kind of change has occurred? OA It's a chemical reaction because a new substance has formed. O B. It's a physical change because the water and the salt kept their original properties. O C. It's an example of synthesis because two substances reacted to give one substance. O D. It's a chemical reaction because the water now tastes salty but originally did not.
HELLLPPPP PLEASEEEEE!!!!!
​
Chemistry
2 answers:
harkovskaia [24]3 years ago
6 0
: B is your answer
:
Neko [114]3 years ago
4 0

Answer:

B. It's a physical change because the water and the salt kept their original properties.

Explanation:

Even though it tastes like salty water, the salt is still present, just in smaller molecules. It can be brought back so it is a physical change.

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A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture co
DaniilM [7]

Answer:

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>

<em></em>

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>

43058 mol air×29g/mol <em>1249 kg air</em>

Percent of oxygen is: \frac{289kg}{1249 kg} =<em>0,231 kg O₂/ kg air</em>

<em></em>

I hope it helps!

4 0
2 years ago
HELP!! BEST ANSWER GETS BRAINLIEST!!
vampirchik [111]
<span> The anwser is D. Radiation, because it can travel thru air.</span>
4 0
3 years ago
Is the salt solution in the Dead<br>Sea dilute, concentrated or<br>saturated? Explain.​
dexar [7]
The waters of the Dead Sea are extremely saline, and, generally, the concentration of salt increases toward the lake's bottom. ... The deep water was saturated with sodium chloride , which precipitated to the bottom.
5 0
2 years ago
0.40 L of HNO3 is titrated to equivalence using 0.18 L of 0.1 M NaOH. What is the concentration of the HNO3 ?
e-lub [12.9K]

This problem is providing us with the volume of nitric acid that is titrated with 0.18 L of 0.1-M sodium hydroxide and asks for the concentration of the acid. At the end, the result turns out to be 0.045M, according to the following.

<h3>Acid-base titrations:</h3>

In chemistry, acid-base titrations allow us to quantify the volume or concentration of an acid or base via the following equation:

M_AV_A=M_BV_B

Where the subscript A stands for the acid and B for the base; which means one can calculate any of the variables there by knowing the other three. This equation is based on the balanced neutralization chemical equation, which takes place between the acid and the base.

Thus, we can write the reaction between NaOH and HNO3 as:

HNO_3+NaOH\rightarrow NaNO_3+H_2O

In such a way, we can solve for the concentration of the acid as shown below:

M_A=\frac{M_BV_B}{V_A} \\\\M_A=\frac{0.1M*0.18L}{0.40L} \\\\M_A=0.045M

Learn more about titration: brainly.com/question/25485091

3 0
2 years ago
Suppose 0.795 g of sodium iodide is dissolved in 100. mL of a 39.0 m M aqueous solution of silver nitrate.
lubasha [3.4K]

Answer:

The final molarity of iodide anion is 0.053 M

Explanation:

<u>Step 1</u>: Data given

Mass of sodium iodide (NaI) = 0.795 grams

Volume of the solution = 100 mL = 0.1 L

Molarity of aqueous solution of silver nitrate (AgNO3) = 39 mM = 0.039M

The molecular mass of sodium iodide is 149.89 g/mol.

<u>Step 2:</u> The balanced equation

AgNO3(aq) + NaI(aq) → AgI(s) + NaNO3(aq)

<u>Step 3: </u>Calculate number of moles of sodium iodide

Moles NaI = mass NaI / Molar mass NaI

Moles NaI = 0.795 grams / 149.89 g/mol

Moles NaI = 0.0053 moles

For 1 mole AgNO3 consumed, we need 1 mole NaI to produce 1 mole AgI and 1 mole NaNO3

The sodium iodide will dissociate as followed:

NaI(aq) → Na+(aq) +  I-(aq)

<u>Step 4</u>: Calculate iodide ions

For 1 mole NaI, we have 1 mole of I-

For 0.0053 moles of NaI we'll have 0.0053 moles I-

<u>Step 5:</u> Calculate molarity of iodide ion

Molarity = moles I- / volume

Molarity I- = 0.0053 moles / 0.1 L

Molarity I- = 0.053 M

The final molarity of iodide anion is 0.053 M

5 0
3 years ago
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