SbF5 mass ratio and atomic ratio

How many moles are there in 45.0 grams of sulfuric acid, H2SO4?

olga2289 [7]

Answer:

the answer is 0.4588162459

Explanation:

1 mole = 0.010195916576195

= 45 ×"

5 0

2 years ago

When 50 ml (50 g) of 1.00 m hcl at 22oc is added to 50 ml (50 g) of 1.00 m naoh at 22oc in a coffee cup calorimeter, the tempera

vitfil [10]

Answer:

$\boxed{\text{2700 J}}$

Explanation:

HCl + NaOH ⟶ NaCl + H₂O

There are two energy flows in this reaction.

Heat of reaction + heat to warm water = 0

q₁ + q₂ = 0

q₁ + mCΔT = 0

Data:

m(HCl) = 50 g

m(NaOH) = 50 g

T₁ = 22 °C

T₂ = 28.87 °C

C = 4.18 J·°C⁻¹g⁻¹

Calculations:

m = 50 + 50 = 100 g

ΔT = 28.87 – 22 = 6.9 °C

q₂ = 100 × 4.18 × 6.9 = 2900 J

q₁ + 2900 = 0

q₁ = -2900 J

The negative sign tells us that the reaction produced heat.

The reaction produced $\boxed{\textbf{2900 J}}$ .

7 0

2 years ago

A converging lens with a focal length of 70.0cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. Th

Setler [38]

Answer: (i)The object is at a distance of 120cm from the lens while the image is at a distance of 160cm from the lens. (ii)The image is Real

Note: This is the complete question; A converging lens with a focal length of 70.0cm forms an image of a 3.20-cm-tall real object that is to the left of the lens. The image is 4.50cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?

Explanation:

The required formulas are;

1. lens formula given by, 1/s' + 1/s = 1/f

2. magnification = h'/h = s'/s

where h' is image height, h is object height, s' is image distance, s is object distance, f is focal length of lens.

h' = 4.50cm, h = 3.20cm, f = 70cm ( f of a converging lens is positive)

solving for s' and s in eqn (2)

4.50/3.20=s'/s

1.4=s'/s

s'=1.4s

to find the values of s' and s, we use equation (1) and substitute s' = 1.4s

1/1.4s + 1/s = 1/70