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ozzi
3 years ago
15

What is the formula for a hydrate that contains 5.99 g CuSO4 and 5.4 g H2O?

Chemistry
1 answer:
adoni [48]3 years ago
8 0

A hydrate is a chemical that has water molecules loosely bonded to it. The water molecules are not ... You will be using the hydrate CuSO4 . ?H2O. Sample Calculation-. An empty crucible has a mass of 12.770 grams.

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How does a battery become depleted or out of charge? * -the number of protons and electrons at both terminals are not equal -the
kompoz [17]

Answer:

the number of protons and neutrons at both terminals are equal

Explanation:

When the number of positive charge and negative charge are both equally the terminal, it becomes neutral and out of charge, because first it undergo enough chemical reaction and there is no remaining tendency for positive and negative charges to get separated. When this tendency dies, the battery also will run out of charge.

3 0
3 years ago
Huge amount of energy is being produced in the core of a hot star due to
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5 0
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C. 840 D. 2053...........
faust18 [17]
They have hydrogens numbers wrong its 1.008 not 1.01
4 0
3 years ago
How many moles of co2 are produced when 5.20 mol of ethane are burned in an excess of oxygen?
galben [10]
<span>2C2H6 + 7O2 = 4CO2 + 6H2O
</span>
According to the equation of the reaction of ethane combustion, ethane and carbon dioxide have following stoichiometric ratio:

n(C2H6) : n(CO2) = 1 : 2

n(CO2) = 2 x n(C2H6) 

n(CO2) = 2 x 5.2 = 10.4 mole of CO2 is formed


6 0
3 years ago
The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically. The following data was obtained: t/min 0
o-na [289]

Answer:

1) The order of the reaction is of FIRST ORDER

2)   Rate constant k = 5.667 × 10 ⁻⁴

Explanation:

From the given information:

The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.

liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.

The following data was obtained:

t/min                    0         10         20          30             40          ∞

conc B/(mol/L)    0       0.089    0.153     0.200       0.230    0.312

For  a first order reaction:

K = \dfrac{1}{t} \ In ( \dfrac{C_{\infty} - C_o}{C_{\infty} - C_t})

where :

K = proportionality  constant or the rate constant for the specific reaction rate

t = time of reaction

C_o = initial concentration at time t

C _{\infty} = final concentration at time t

C_t = concentration at time t

To start with the value of t when t = 10 mins

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 - 0}{0.312 - 0.089})

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 }{0.223})

K_1 =0.03358 \  min^{-1}

K_1 \simeq 0.034 \  min^{-1}

When t = 20

K_2= \dfrac{1}{20} \ In ( \dfrac{0.312 - 0}{0.312 - 0.153})

K_2= 0.05 \times  \ In ( 1.9623)

K_2=0.03371 \ min^{-1}

K_2 \simeq 0.034 \ min^{-1}

When t = 30

K_3= \dfrac{1}{30} \ In ( \dfrac{0.312 - 0}{0.312 - 0.200})

K_3= 0.0333 \times  \ In ( \dfrac{0.312}{0.112})

K_3= 0.0333 \times  \ 1.0245

K_3 = 0.03412 \ min^{-1}

K_3 = 0.034 \ min^{-1}

When t = 40

K_4= \dfrac{1}{40} \ In ( \dfrac{0.312 - 0}{0.312 - 0.230})

K_4=0.025 \times  \ In ( \dfrac{0.312}{0.082})

K_4=0.025 \times  \ In ( 3.8048)

K_4=0.03340 \ min^{-1}

We can see that at the different time rates, the rate constant of k_1, k_2, k_3, and k_4 all have similar constant values

As such :

Rate constant k = 0.034 min⁻¹

Converting it to seconds ; we have :

60 seconds = 1 min

∴

0.034 min⁻¹ =(0.034/60) seconds

= 5.667 × 10 ⁻⁴ seconds

Rate constant k = 5.667 × 10 ⁻⁴

4 0
3 years ago
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