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Advocard [28]
3 years ago
13

Question:

Chemistry
1 answer:
Elina [12.6K]3 years ago
8 0

H2SO4 + ZN ------- ZNSO4+ H2

(SO4)²The sulphate salt is formed......

Hope it helps

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In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of Na2CO3 was quickly spread on the area
storchak [24]

First we must write a balanced chemical equation for this reaction

Na_2CO_3 _(_a_q_)+ 2HNO_3_(_a_q_) \implies 2NaNO_3_(_s_) + CO_2_(_g_) + H_2O_(_l_)

The mole ratio for the reaction between HNO_3 and Na_2CO_3 is 1:2. This means 1 moles of Na_2CO_3 will neutralize 2 moles HNO_3. Now we find the moles of each reactant based on the mass and molar mass.

2500g HNO_3 \times \frac{mol}{63.01g\ HNO_3} = 39.67 mol\ HNO_3

2000g\ Na_2CO_3 \times \frac{mol}{105.99g \ Na_2CO_3} = 18.87 mol\ Na_2CO_3

\frac{18.87 mol Na_2CO_3}{39.67\ HNO_3} = \frac{1 molNa_2CO_3}{2 mol HNO_3}

The Na_2CO_3 was enough to neutralize the acid because 18.87:39.67 is the same as 1:2 mol ratio.

4 0
3 years ago
A sample of an unknown gas occupies 5.51 L at 1.31 atm. What pressure would thisgas exert in a 0.520 L container if the temperat
weeeeeb [17]

Answer:

P₂ = 13.9 atm (3 sig. figs.)

Explanation:

The pressure (P), Volume (V) relationship with Temperature (T) & mass (n) held constant is an inverse proportionality. That is Boyles Law ...

P ∝ 1/V => P = k/V => k = P·V

For two pressure-volume conditions, the proportionality constant (k) remains constant where k₁ = k₂ and P₁·V₁ = P₂·V₂ => P₂ = P₁·V₁/V₂

Given:

P₁ = 1.31 atm.

V₁ = 5.51 L

P₂ = ?

V₂ = 0.520 L

V₂ = (1.31 atm)(5.51L)/(0.520L) = 13.88096154 atm (calc. ans.) = 13.9 atm (3 sig. figs.)

5 0
3 years ago
a) A table tennis ball has been hit is about 13.16 times as fast as the speed for the swimming. What is the speed of the table t
Alex

x=swimming

The answer is X*13.16

7 0
3 years ago
3. Burns from boiling water can be severe, caused by the transfer of energy from the boiling water to the
Free_Kalibri [48]

Answer:

Q = 3937.56  J

Explanation:

Heat transferred due to change in temperature is given by :

Q=mc\Delta T

c is the specific heat of water, c=4.18 J/g-°C

We have, m = 15 g, T_i=100^{\circ} C\ \text{and}\ T_f=37.2^{\circ} C

So,

Q=15\times 4.18\times (37.2-100)\\Q=-3937.56\ J

Hence, 3937.56  J of heat is transferred.

8 0
3 years ago
The value of Henry's law constant for oxygen in water at 242C is 1.66 x 1 ox M/torr. b. c. Calculate the solubility of oxygen in
DochEvi [55]
25ca an the oxygen gas is 2 atom ur mum
4 0
3 years ago
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