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nadya68 [22]
2 years ago
5

What do we mean by basic MgO?

Chemistry
1 answer:
aliina [53]2 years ago
5 0

Answer:

Magnesium oxide is a simple basic oxide, because it contains oxide ions. It reacts with water to form magnesium hydroxide which is a base.

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Question 5
bagirrra123 [75]

Answer:

true

Explanation:

any object that is larger will take a longer time to do things, the same applies to heat levels. The lower the heat, the longer it takes, the higher the heat, the shorter it takes. So if an iceberg is large, it will need a higher heat, whereas an ice cube is really small and doesn't need that much heat to melt.

3 0
3 years ago
A balloon occupies a volume of 2.00 l at 40.0oc. how much volume will it occupy at 30.0oc?
hodyreva [135]
<span>Charles' law says "at a constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature".

V </span>α T

Where V is the volume and T is the temperature in Kelvin of the gas. We can use this for two situations as,
V₁/T₁ = V₂/T₂

V₁ = 2.00 L
T₁ = 40.0 ⁰C = 313 K
V₂ = ?
T₂ = 30.0 ⁰C = 303 K

By applying the formula,
2.00 L / 313 K = V₂ / 303 K                   
                   V₂ = (2.00 L / 313 K) x 303 K
                   V₂ = 1.94 L

Hence, the volume of the balloon at 30.0 ⁰C  is 1.94 L
7 0
3 years ago
The reaction for photosynthesis producing glucose sugar and oxygen gas is:
uranmaximum [27]

Answer:

a. 1.12 L

Explanation:

Step 1: Write the balanced equation for the photosynthesis

6 CO₂(g) + 6 H₂O(l) ⇒ C₆H₁₂O₆(s) + 6 O₂(g)

Step 2: Calculate the moles corresponding to 2.20 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

2.20 g × 1 mol/44.01 g = 0.0500 mol

Step 3: Calculate the moles of O₂ produced

The molar ratio of CO₂ to O₂ is 6:6. The moles of O₂ produced are 6/6 × 0.0500 mol = 0.0500 mol

Step 4: Calculate the volume occupied by 0.0500 moles of O₂ at STP

At STP, 1 mole of O₂ occupies 22.4 L.

0.0500 mol × 22.4 L/1 mol = 1.12 L

4 0
2 years ago
All molecules have the same kinetic energy and hence the same speed.<br><br> O True<br><br> O False
topjm [15]
False...their speed differs
4 0
2 years ago
The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r
rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

    = 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the  rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

K=Ae^{\frac{-E}{RT}}

If K_1=Ae^{\frac{-E_1}{RT}} -------equation 1     &

K_2=Ae^{\frac{-E_2}{RT}} -------equation 2

\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}

E_1= E_2-RT*In(\frac{K_1}{K_2})

E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})

E_1 = 28957.39292  J/mol

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0
2 years ago
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