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3 years ago

Which conditions must be met for two triangles to be similar

1 answer:

7 0

The included angle, i.e. If two pairs of sides of two triangles are in proportion, and the included angles are equal, then the triangles are similar.

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Someone help me plz :))) brainliest

SOVA2 [1]

Answer:

Option 2

Explanation:

3 0

3 years ago

Una muestra de 258.4 g de etanol (C2H5OH) se quemó en una bomba calorimétrica

Ilia_Sergeevich [38]

Answer:

Explanation:

Unclear question.

I infer you want a clear rendering, which reads;

A 258.4 g sample of ethanol (C2H5OH) was burned in a calorimetric pump using a Dewar glass. As a consequence, the water temperature rose to 4.20 ° C.

If the heat capacity of the water and the surrounding glass was 10.4 kJ / ° C, calculate the heat of combustion of one mole of ethanol.

4 0

3 years ago

What are two ways in which white blood cells fight pathgons that have entered the body

crimeas [40]

Answer:

Answer c i believe

Explanation:

3 0

3 years ago

Write one common thing between condensation and hydrolysis ?

serious [3.7K]

Answer:

The common thing is the compound water

Explanation:

in condensation h2O is expelled while in hydrolysis water is used or added

8 0

2 years ago

A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci

Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

$C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}$

The ICE table is then shown as:

$C_3H_6ClCO_2H_{(aq)} \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ + \ \ \ \ H^+_{(aq)}$

Initial (M) 1.8 0 0

Change (M) - x + x + x

Equilibrium (M) (1.8 -x) x x

$K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}$

where ;

$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}$

Since the value for $K_a$ is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

$3.02*10^{-5} *(1.8) = {(x)(x)}$

$5.436*10^{-5}= {(x^2)$

$x = \sqrt{5.436*10^{-5}}$

$x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M$

Dissociated form of 4-chlorobutanoic acid = $C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M$

Percentage dissociated = $\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100$

Percentage dissociated = $\frac{7.3729*10^{-3}}{1.8 }*100$

Percentage dissociated = 0.4096

Percentage dissociated = 0.41% (to two significant digits)

3 0

3 years ago

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