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3 years ago

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If a copper acetate hydrate, Cux(C2H3O2)yzH20 compound is found to contain 31.82% copper, 59.14% acetate, and the remainder is w

ater. What is the empirical formula?

1 answer:

Aloiza [94]3 years ago

4 0

The Subscripts are 1, 1, 2 Cu(C2H3O2) 2H2O PLZ mark brainleist

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Concrete is formed of sand. Is it incorrect to suggest that some houses are built using concrete blocks?

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2 years ago

What is the decay mode of radium-226?

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Answer:

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3 years ago

A hot lump of 39.9 g of iron at an initial temperature of 78.1 °C is placed in 50.0 mL H 2 O initially at 25.0 °C and allowed to

Drupady [299]

Answer : The final temperature of the mixture is $29.6^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron = $0.499J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of iron = 39.9 g

$m_2$ = mass of water = $Density\times Volume=1g/mL\times 50.0mL=50.0g$

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of iron = $78.1^oC$

$T_2$ = initial temperature of water = $25.0^oC$

Now put all the given values in the above formula, we get

$(39.9g)\times (0.499J/g^oC)\times (T_f-78.1)^oC=-(50.0g)\times 4.18J/g^oC\times (T_f-25.0)^oC$

$T_f=29.6^oC$

Therefore, the final temperature of the mixture is $29.6^oC$

8 0

3 years ago

Help me if you can please help me please

Dovator [93]

Answer: with? i can help:))

Explanation:

hmu love

8 0

3 years ago

Read 2 more answers

What is the Ka of a weak acid (HA) if the initial concentration of weak acid is 4.5 x 10-4 M and the pH is 6.87? (pick one)

rewona [7]

Answer:

Ka = $4.04 \times 10^{-11}$

Explanation:

Initial concentration of weak acid = $4.5 \times 10^{-4}\ M$

pH = 6.87

$pH = -log[H^+]$

$[H^+]=10^{-pH}$

$[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M$

HA dissociated as:

$HA \leftrightharpoons H^+ + A^{-}$

(0.00045 - x) x x

[HA] at equilibrium = (0.00045 - x) M

x = $1.35 \times 10^{-7}\ M$

$Ka = \frac{[H^+][A^{-}]}{[HA]}$

$Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}$

0.000000135 <<< 0.00045

$Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}$

5 0

3 years ago