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3 years ago

All organisms have the same number of cellsA. trueB. false

Chemistry

2 answers:

denis-greek [22]3 years ago

8 0

The answer is B because not all organisms have the same number of cells

MakcuM [25]3 years ago

4 0

B. false is the answer of this question

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You have a 15.0 gram sample of gold at 20.0°C. How much heat does it take to raise the temperature to 100.0°C?

Nadusha1986 [10]

Answer:

=154.8 J

Explanation:

The rise in temperature is contributed by the change in temperature.

Change in enthalpy = MC∅, where M is the mass of the substance, C is the specific heat capacity and ∅ is the change in temperature.

Change in temperature = 100.0°C-20.0°C=80°C

ΔH=MC∅

The specific heat capacity of gold= 0.129 J/g°C

ΔH= 15.0g×0.129J/g°C×80°C

=154.8 J

7 0

3 years ago

Find the volume in milliliters of 2.00 mol of an ideal gas at 36°C and a pressure of 1120 torr.

hram777 [196]

Answer:

V = 34430 mL

Explanation:

Given data:

Volume in mL = ?

Number of moles of gas = 2.00 mol

Temperature = 36°C (36+273= 309K)

Pressure of gas = 1120 torr

Solution:

Formula:

PV = nRT

V = nRT/P

V = 2.00 mol ×62.4 torr • L/mol · K × 309K / 1120 torr

V = 38563.2 torr • L / 1120 torr

V = 34.43 L

L to mL

34.43 L ×1000 mL / 1 L

34430 mL

5 0

2 years ago

Three quantum numbers for an electron in a hydrogen atom in a certain state are

OLga [1]

It is option A 4s cause n=4, l=1 this is 4s orbital

6 0

2 years ago

At the equilibrium point in the decomposition of phosphorus pentachloride to chlorine and phosphorus trichloride, the following

Hitman42 [59]

Answer: $K_{eq}$ for the reaction is 5.55

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

At eqm. conc. (0.010) M (0.15) M (0.37) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[Cl_2]\times [PCl_3]}{[PCl_5]}$

Now put all the given values in this expression, we get :

$K_c=\frac{(0.37)\times (0.15)}{(0.010)}$

$K_c=5.55$

Thus the $K_{eq}$ for the reaction is 5.55

5 0

3 years ago

Combustion of 9.511 grams of c4h10 will yield ____ grams of CO2

Flauer [41]

Answer:

$\boxed{28.81}$

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 58.12 44.01

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

m/g: 9.511

1. Moles of C₄H₁₀

$\text{Moles of C$_{4}$H$_{10} $} = \text{ 9.511 g C$_{4}$H$_{10} $} \times \dfrac{\text{1 mol C$_{4}$H$_{10} $}}{\text{ 58.12 g C$_{4}$H$_{10} $}} = \text{0.1636 mol C$_{4}$H$_{10}$}$

2. Moles of CO₂

The molar ratio is 8 mol CO₂:2 mol C₄H₁₀

$\text{Moles of CO}_{2} =\text{0.1636 mol C$_{4}$H$_{10} $} \times \dfrac{\text{8 mol CO}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \text{0.6546 mol CO}_{2}$

3. Mass of CO₂

$\text{Mass of CO}_{2} = \text{0.6546 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO}_{2}} = \textbf{28.81 g CO}_{2}\\\\\text{The combustion will form $\boxed{\textbf{28.81 g CO}_{2}}$}$

8 0

2 years ago

All organisms have the same number of cellsA. trueB. false​

All organisms have the same number of cellsA. trueB. false