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Dovator [93]
3 years ago
15

can you help me with #7? #7 says "A substance enters into and is altered through the course of a chemical change. it is called a

___." pick one of these choices. The lines that are next to the words are already picked from #s 1, 2, 3, 4, 5,& 6. do not pick more than one answer pick one answer only. tap the image so you could see the full image. (What I'm talking about is science.)

Chemistry
2 answers:
ad-work [718]3 years ago
8 0
It's reactant. That's what I'm pretty sure it js
Zigmanuir [339]3 years ago
3 0
Reactant hope this helps
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The downhill flow of soil and rock resulting from gravity best describes mass movement.

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Which statement about the gas laws is correct?
Elena-2011 [213]

Answer:

Gay-Lussac's law states that pressure and temperature are directly proportional

Explanation:

Gay-Lussac's law states that pressure and temperature are directly proportional. This always occurs if the volume keeps in constant.

n and V are not directly proportional, they are the same.

At Charles Gay Lussac's law

V1 = V2

n1 = n2

T1 < T2

P1 < P2

P1 / T1 = P2 / T2

If the pressure is contant:

V1 / T1 = V2 /T2

4 0
3 years ago
50cm3 of 1 mol/dm3 HCl at 30°C was mixed with 50cm3 of 1mol/dm3 NaOH at 30°C in a styrofoam calorimeter. The temperature of the
trapecia [35]

Answer:

-21 kJ·mol⁻¹  

Explanation:

Data:

                    H₃O⁺ +  OH⁻ ⟶ 2H₂O

       V/mL:    50         50  

c/mol·dm⁻³:   1.0         1.0

     

ΔT = 4.5 °C  

       C = 4.184 J·°C⁻¹g⁻¹

C_cal = 50 J·°C⁻¹

Calculations:

(a) Moles of acid

\text{Moles of acid} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}\\\\\text{Moles of base} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}

So, we have 0.050 mol of reaction

(b) Volume of solution

V = 50 dm³ + 50 dm³ = 100 dm³

(c) Mass of solution

\text{Mass of solution} = \text{100 dm}^{3} \times \dfrac{\text{1.00 g}}{\text{1 dm}^{3}} = \text{100 g}

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

q₁ + q₂ + q₃ = 0

     nΔH   +         mCΔT       + C_calΔT = 0

0.050ΔH + 100×4.184×4.5 +   50×4.5  = 0

0.050ΔH +          1883        +      225    = 0

                                  0.050ΔH + 2108 = 0

                                              0.050ΔH = -2108

                                                        ΔH = -2108/0.0500

                                                              = -42 000 J/mol

                                                              = -42 kJ/mol

This is the heat of reaction for the formation of 2 mol of water

The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.

5 0
3 years ago
A gas occupies a 1.5 l container at 25 degrees
musickatia [10]
The answer is:

683.8 cm^3
7 0
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