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2 years ago

What are the lengths of the diagonals of the kite?

Chemistry

1 answer:

umka21 [38]2 years ago

8 0

The answer ( 13 and 8 )

x²=5²+12²

x²=25+144

x²=169

x=13

x²=5²+6²

x²=25+36

x²=61

x=7.8

x=8

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MAVERICK [17]

Answer:

Explanation:

7 0

2 years ago

The mole fraction of CO2 in a certain solution with H2O as the solvent is 3.6 × 10−4. What is the approximate molality of CO2 in

nataly862011 [7]

Answer: C) 0.020 m

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

$Molality=\frac{n\times 1000}{W_s}$

where,

n = moles of solute

$W_s$ = weight of solvent in g

Mole fraction of $CO_2$ is = $3.6\times 10^{-4}$ i.e. $3.6\times 10^{-4}$ moles of $CO_2$ is present in 1 mole of solution.

Moles of solute $(CO_2)$ = $3.6\times 10^{-4}$

moles of solvent (water) = 1 - $3.6\times 10^{-4}$ = 0.99

weight of solvent = $moles\times {\text {Molar mass}}=0.99\times 18=17.82g$

Molality = $\frac{3.6\times 10^{-4}\times 1000}{17.82g}=0.020$

Thus approximate molality of $CO_2$ in this solution is 0.020 m

5 0

2 years ago

Which ONE of the following is an oxidation–reduction reaction? A) PbCO3(s) + 2 HNO3(aq) ––––> Pb(NO3)2(aq) + CO2(g) + H2O(l)

sveta [45]

Answer:

E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g)

Explanation:

Which ONE of the following is an oxidation–reduction reaction?

A) PbCO₃(s) + 2 HNO₃(aq) ⇒ Pb(NO₃)₂(aq) + CO₂(g) + H₂O(l). NO. All the elements keep the same oxidation numbers.

B) Na₂O(s) + H₂O(l) ⇒ 2 NaOH(aq). NO. All the elements keep the same oxidation numbers.

C) SO₃(g) + H₂O(l) ⇒ H₂SO₄(aq). NO. All the elements keep the same oxidation numbers.

D) CO₂(g) + H₂O(l) ⇒ H₂CO₃(aq). NO. All the elements keep the same oxidation numbers.

E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g). YES. <u>C is reduced</u> and <u>H is oxidized</u>.

8 0

2 years ago

Write a nuclear equation for the beta decay of Promethium-165

padilas [110]

Answer:

see your answer in pic

Explanation:

4 0

2 years ago

A balloon at 25°C has 30 L. What will the balloon's volume at 35°C?

anygoal [31]

The balloon's volume at 35°C : V₂=31.01 L

<h3>Further explanation</h3>

Given

T₁ = 25°C+273 = 298 K

V₁ = 30 L

T₂ = 35 °C + 273 = 308 K

Required

The new volume (V₂)

Solution

Charles's Law

When the gas pressure is kept constant, the gas volume is proportional to the temperature

$\tt \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

Input the value :

V₂=(V₁.T₂)/T₁

V₂=(30 x 308)/298

V₂=31.01 L

8 0

2 years ago