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Ludmilka [50]
2 years ago
13

What can lead to being curious and also lead to fascinating scientific investigations? a. Questions b. Answers c. Qualitative Me

asurements d. Quantitative Measurements
Chemistry
1 answer:
Ira Lisetskai [31]2 years ago
8 0

Answer:

Questions

Explanation:

The word "science" was obtained from the Latin word "Scire" which means "to know". Hence science is a systematic investigation into the nature of the universe.

Scientific investigations begin with asking questions about the universe. It is this curiosity to know about occurrences in the universe as well as how to solve problems in the universe that leads to fascinating scientific investigations and discoveries.

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For an object to sink
AfilCa [17]

Answer:

For and object to sink, it must have more density than the liquid in which it is placed. For example, if you have a glass of water and a metal spoon, the spoon will sink because it is both heavier than the water, therefore having more density.

5 0
3 years ago
How much heat is required, in calories, to raise the temperature of 57.8 g of silver from
Stella [2.4K]

Answer:

87.3 calories of heat is required.

Explanation:

Heat = mcΔT

m= mass, c = specific heat of silver, T = temperature

H= 57.8 g * 0.057 cal/g°C * ( 43.5 - 17 °C)

H = 57.8 * 0.057 * 26.5

H = 87.3069 cal.

The heat required to raise the temperature of 57.8 g of silver from 17 °C to 43.5 °C is 87.3 calories.

8 0
2 years ago
If a reaction is reversible what are the relative amounts of reactant and product
Alina [70]
<span>If a reaction is reversible, then it will attain the phase of Equilibrium and at that phase, the Amount of Reactants and Products would be:  Equal

Hope this helps!</span>
3 0
3 years ago
Read 2 more answers
Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 5.90 moles of magnesium perchlorate, Mg(ClO4)2.
Lady bird [3.3K]

Answer:

5.90, 11.8, 47.2

Explanation:

Let’s remove the parentheses and write the formula as MgCl₂O₈.

We see that 1 mol Mg(ClO₄)₂ contains 1 mol Mg atoms, 2 mol Cl atoms, and 8 mol O atoms.

∴ \text{Moles of Mg atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{1 mol Mg atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{5.90 mol Mg atoms}  

\text{Moles of Cl atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{2 mol Cl atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{11.8 mol Cl atoms}

\text{Moles of O atoms} = \text{5.90 mol Mg(ClO}_{4})_{2} \times \frac{\text{8 mol O atoms}} {\text{1 mol Mg(ClO}_{4})_{2}} = \text{47.2 mol O atoms}

∴ Mg, Cl, O = 5.90, 11.8, 47.2

4 0
3 years ago
The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A
liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

p^{OH}=14-56.17

      =8.823

The volume of the NH_{3} = 40.00 ml  

convert into the liter= 0.040L

The value of the concentrated NH_{3} =0.10 M

The volume of the NH_{4}Cl= 50.00 ml

convert into the liter= 0.050L

 The value of concentrated NH_{4}Cl= 0.10 M

The volume of the H_{2}So_{4}= 30 ml

convert into the liter= 0.030L  

The value of concentrated H_2So_4=0.05 M

Calculating total volume=(0.40+0.050+0.030)

                                       =0.120 L

calculating the new concentrated value of NH_3 = \frac{0.10\times 0.040}{0.120}= 0.33 \ M

calculating the new concentrated value of NH_4Cl= \frac{0.050\times 0.10}{0.120}= 0.04166 \ Mcalculating the new concentrated value of H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M when 1 mol H_2So_4 produced 2 mols H^{+} so, 0.0125 in H_2So_4produced:

=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}

create the ICE table:    

NH_3    \ \ \ \ \ \ \ \     + H^{+}  \ \ \ \ \ \ \longrightarrow NH_4^{+}                    

I (m)       0.033(m)            0.025                       0.04166

C            -0.025                 -0.025                       + 0.025  

E            8.3\times 10^{-3}     0                    0.0667

now calculating pH:

when ph= 8.83:

P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769

5 0
3 years ago
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