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grigory [225]
3 years ago
15

Which of the following is an example of a hypothesis?

Chemistry
2 answers:
Marrrta [24]3 years ago
7 0

Answer: A). a prediction that a tire's air pressure will increase if the tire's temperature is raised

Explanation:

A hypothesis is a proposed explanation for a phenomenon. For a hypothesis to be a scientific hypothesis, the scientific method requires that one can test it. Scientists generally base scientific hypotheses on previous observations that cannot satisfactorily be explained with the available scientific theories.

A hypothesis can be define as the presumed or predicted statement for a natural phenomena which is required to be testified by using suitable experimental methodology or observatory approaches.

A). a prediction that a tire's air pressure will increase if the tire's temperature is raised. is the correct answer this is because of the fact that the statement involves a prediction that can be proved by using suitable experimental methodology.

nevsk [136]3 years ago
5 0
The answer would be A.
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3 years ago
In a closed system containing three different gases as shown, X2 + Y2 = 2XY
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3 years ago
A detailed description please...
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3 years ago
Use the image to rank the colors from the color that will refract the least to the color that will refract the most.
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5 0
3 years ago
Read 2 more answers
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
Brums [2.3K]

Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

7 0
3 years ago
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