Please help!!

Suppose you wish to apply SSA to a triangle, in order to find an angle

emmasim [6.3K]

First of all there nothing exists something like SSA congruence criterion, It is SAS Congruence where the sides of a Triangle and the angle between them is exactly same to the other two sides and angle between them of another triangle.

So the correct option is D.

Explanation:

If the Triangle has two sides equal, so the two angles there are equal. Also the other angle be some other definite angle. So

45+x+x=180(Considering definite angle to be 45)
2x=135
x=67.5

So there is only one solution to the triangle.

8 0

2 years ago

For which of the following reactions is S° > 0. Choose all that apply. N2(g) + 3H2(g) 2NH3(g) 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2

Lostsunrise [7]

Answer: $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

$NH_4I(s)\rightarrow NH_3(g)+HI(g)$

$2H_2O(g)+2SO_2(g)\rightarrow 2H_2S(g)+3O_2(g)$

Explanation:

Entropy is the measure of randomness or disorder of a system.

A system has positive value of entropy if the disorder increases and a system has negative value of entropy if the disorder decreases.

1. $N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

As 4 moles of gaseous reactants are changing to 2 moles of gaseous products, the randomness is decreasing and the entropy is negative

2. $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

As 9 moles of gaseous reactants are changing to 10 moles of gaseous products, the randomness is increasing and the entropy is positive.

3. $NH_4I(s)\rightarrow NH_3(g)+HI(g)$

As 1 mole of solid reactants is changing to 2 moles of gaseous products, the randomness is increasing and the entropy is positive.

4. $2H_2O(g)+2SO_2(g)\rightarrow 2H_2S(g)+3O_2(g)$

As 4 moles of gaseous reactants is changing to 5 moles of gaseous products, the randomness is increasing and the entropy is positive

5. $2NO(g)+2H_2(g)\rightarrow N_2(g)+2H_2O(l)$

As 4 moles of gaseous reactants is changing to 1 moles of gaseous products, the randomness is decreasing and the entropy is negative.

5 0

3 years ago

A 55.0 mL aliquot of a 1.50 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a

balu736 [363]

Answer:

0.14 M

Explanation:

To determinate the concentration of a new solution, we can use the equation below:

C1xV1 = C2xV2

Where C is the concentration, and V the volume, 1 represents the initial solution, and 2 the final one. So, first, the initial concentration is 1.50 M, the initial volume is 55.0 mL and the final volume is 278 mL

1.50x55.0 = C2x278

C2 = 0.30 M

The portion of 139 mL will be the same concentration because it wasn't diluted or evaporated. The final volume will be the volume of the initial solution plus the volume of water added, V2 = 139 + 155 = 294 mL

Then,

0.30x139 = C2x294

C2 = 0.14 M

4 0

2 years ago

Read 2 more answers

Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow

andreyandreev [35.5K]

Explanation:

(a) The given data is as follows.

Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )