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Vikki [24]
2 years ago
13

Where do the sun rays hit Earth's surface nearly at right angles ​

Chemistry
1 answer:
hoa [83]2 years ago
8 0

Answer:

The Sun Ray's hit earths surface at Earths Equator

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Phosphorus that is stored in organisms
Mumz [18]

Explanation:

B. Recycles slowly

And it also depends on the organism

3 0
3 years ago
How many moles are in 1.05 g of gold (Au)?
Wittaler [7]

Answer:

0.005 mol

Explanation:

Moles is denoted by given mass divided by the molecular mass ,  

Hence ,  

n = w / m

n = moles ,  

w = given mass ,  

m = molecular mass .

From the question ,

w = given mass of Gold = 1.05 g ,

m = molecular mass of Gold = 197 g/mol

<u>Hence , moles can be calculated as -</u>

n = w / m = 1.05 g / 197 g/mol = 0.005 mol

7 0
3 years ago
What is the ratio of [a–]/[ha] at ph 2.75? the pka of formic acid (methanoic acid, h–cooh) is 3.75?
Varvara68 [4.7K]

The dissociation of formic acid is:

HCOOH \rightleftharpoons HCOO^{-} + H^{+}

The acid dissociation constant of formic acid, k_a is:

k_a = \frac{[HCOO^{-}]  [H^{+}]}{HCOOH}

Rearranging the equation:

\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}

pH = 2.75

pH = -log[H^{+}]

[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}

pk_a = 3.75

k_a = 10^{-3.75} = 1.78\times 10^{-4}

Substituting the values in the equation:

\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}

\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}

Hence, the ratio is \frac{1}{10}.

4 0
3 years ago
A balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm,
choli [55]

Answer:

3.676 L.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R  is the general gas constant,

T is the temperature of the gas in K.

If n and P are constant, and have different values of V and T:

(V₁T₂) = (V₂T₁)

Knowing that:

V₁ = 3.5 L, T₁ = 25°C + 273 = 298 K,

V₂ = ??? L, T₂ = 40°C + 273 = 313 K,

Applying in the above equation

(V₁T₂) = (V₂T₁)

∴ V₂ = (V₁T₂)/(T₁) = (3.5 L)(313 K)/(298 K) = 3.676 L.

7 0
3 years ago
53. A firefighter of mass 80 kg slides down a vertical pole
Zielflug [23.3K]

Answer:

Answer is given below:

Explanation:

<em>Given Data:</em>

mass = 80kg

acceleration = 4 ms

force = 800N

<em>Find out:</em>

friction = ?

<em>Formula</em><em>:</em>

F-friction = weight - f-net

<em>Solution:</em>

weight = (80)(10)

           = 800 N

F-net = ma =(80)(4) = 320N

F-friction = weight - F-net

               =800 N - 320N

               =480N

<em>Answer</em> :

Friction = 480 N

   

6 0
3 years ago
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