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Paul [167]
3 years ago
12

Jimmy mixes 2 chemicals in the lab. The chemicals change color and they get hot. The rise in temperature

Chemistry
1 answer:
gladu [14]3 years ago
3 0

Answer:

A chemical reaction.

Explanation:

A change in temperature is evidence of a chemical reaction.

Also: They are chemicals...

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True or False : Not all metals are ferromagnetic-only the ones with iron,<br> nickel, or cobalt. *
Triss [41]
Answer: True
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5 0
2 years ago
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If 4.168 kJ of heat is added to a calorimeter containing 75.40 g of water, the temperature of the water and the calorimeter incr
Alinara [238K]

Answer:

The value of  the heat capacity of the Calorimeter  C_c = 54.4 \frac{J}{c}

Explanation:

Given data

Heat added Q = 4.168 KJ = 4168 J

Mass of water m_w = 75.40 gm

Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c

From the given condition

Q = m_w C_w ΔT + C_c ΔT

Put all the values in above equation we get

4168 = 75.70 × 4.18 × 11.24 +  C_c × 11.24

611.37 =  C_c × 11.24

C_c = 54.4 \frac{J}{c}

This is the value of  the heat capacity of the Calorimeter.

7 0
3 years ago
How does increasing the temperature affect a reaction?
Luden [163]

Answer:A. An increase in temperature increases the reaction rate.

Explanation:because of the disproportionately large increase in the number of high energy collisions. It is only these collisions (possessing at least the activation energy for the reaction) which result in a reaction.

4 0
3 years ago
As a pendulum swings from its highest to its lowest position along an arc, what happens to its kinetic energy and potential ener
lara31 [8.8K]

C.

Explanation:

As a pendulum swings from its highest to its lowest position along an arc, what happens to its kinetic energy and potential energy? The potential energy decreases while the kinetic energy increases.

4 0
2 years ago
Equal molar quantities of Ca2 and EDTA (H4Y) are added to make a 0.010 M solution of CaY2- at pH 10. The formation constant for
abruzzese [7]

Answer:

the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

Explanation:

Given the data in the question;

Ca^{2+ + y^{4- ⇄  CaY^{2-

Formation constant Kf

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) = 5.0 × 10¹⁰

Now,

[y^{4-] = \alpha _4CH_4Y; ∝₄ = 0.35

so the equilibrium is;

Ca^{2+ + H_4Y ⇄  CaY^{2- + 4H⁺

Given that; CH_4Y = Ca^{2+     { 1 mol Ca^{2+  reacts with 1 mol H_4Y  }

so at equilibrium, CH_4Y = Ca^{2+ = x

∴

Ca^{2+ + y^{4- ⇄  CaY^{2-

x        + x         0.010-x

since Kf is high, them x will be small so, 0.010-x is approximately 0.010

so;

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) =  CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ 0.010 / ( [x][ 0.35 × x] )  = 5.0 × 10¹⁰

⇒ 0.010 / 0.35x²  = 5.0 × 10¹⁰

⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 5.7142857 × 10⁻¹³

⇒ x = √(5.7142857 × 10⁻¹³)

⇒ x = 7.559 × 10⁻⁷

Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

8 0
3 years ago
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