Help pls i need help and i’m tired i took 15 test today-

Which situations make use of interactive multimedia and which do not?

xxMikexx [17]

Fred is using interactive for sure. It requires a response from you, not just looking, watching, or listening.

4 0

4 years ago

Please help me find the final answer

dlinn [17]

Answer:

5. The mass of Na₂CO₃, that will produce 5 g of CO₂ is approximately 12.04 grams of Na₂CO₃

6. The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃ is 693. $\overline{3168}$ grams of N₂

Explanation:

5. The given equation for the formation of carbon dioxide (CO₂) from sodium bicarbonate (Na₂CO₃) is presented as follows;

(Na₂CO₃) + 2HCl → 2NaCl + CO₂ + H₂O

One mole (105.99 g) of Na₂CO₃ produces 1 mole (44.01 g) of CO₂

The mass, 'x' g of Na₂CO₃, that will produce 5 g of CO₂ is given by the law of definite proportions as follows;

$\dfrac{x \ g}{105.99 \ g} = \dfrac{5 \ g}{44.01 \ g}$

$\therefore {x \ g} = \dfrac{5 \ g}{44.01 \ g} \times 105.99 \ g \approx 12.04 \ g$

The mass of Na₂CO₃, that will produce 5 g of CO₂, x ≈ 12.04 g

6. The chemical equation for the reaction is presented as follows;

N₂ + 3H₂ → 2NH₃

Therefore, one mole (28.01 g) of nitrogen gas, (N₂), reacts with three moles (3 × 2.02 g) of hydrogen gas (H₂) to produce 2 moles of ammonia (NH₃)

The mass 'x' grams of nitrogen gas (N₂) that will react completely with150 g of hydrogen (H₂) in the production of NH₃ is given as follows;

$\dfrac{x \ g}{28 .01 \ g} = \dfrac{150 \ g}{3 \times 2.02 \ g} = \dfrac{150 \ g}{6.06 \ g}$

$\therefore \ x \ g= \dfrac{150 \ g}{6.06 \ g} \times 28.01 \ g = 693.\overline {3168} \ g$

The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃, x = 693. $\overline{3168}$ grams

6 0

3 years ago

What is a limiting factor?

alukav5142 [94]

A limiting factor helps an organism outcompete other organisms i think that’s the answer

5 0

3 years ago

A 14.4-gg sample of granite initially at 86.0 ∘C∘C is immersed into 24.0 gg of water initially at 25.0 ∘C∘C. What is the final t

kari74 [83]

Answer:

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

Explanation:

Step 1: Data given

Mass of sample granite = 14.4 grams

Initial temperature = 86.0 °C

Mass of water = 24.0 grams

The initial temperature of water = 25.0 °C

The specific heat of water = 4.18 J/g°C

The specific heat of granite = 0.790 J/g°C

Step 2: Calculate the final temperature

Heat lost = heat gained

Qgranite = - Qwater

Q = m*c*ΔT

m(granite)*c(granite)*ΔT(granite) = -m(water)*c(water)*ΔT(water)

⇒with m(granite) = the mass of granite = 14.4 grams

⇒with c(granite) = The specific heat of granite = 0.790 J/g°C

⇒with ΔT⇒(granite) = the change of temperature of granite = T2 - T1 = T2 - 86.0 °C

⇒with m(water) = the mass of water = 24.0 grams

⇒with c(water) = The specific heat of water = 4.18 J/g°C

⇒with ΔT(water) = the change of temperature of granite = T2 - T1 = T2 -25.0°C

14.4 grams * 0.790 * (T2 - 86.0°C) = -24.0 *4.18 * (T2 - 25.0°C)

11.376T2 - 978.336 = -100.32T2 + 2508

111.696 T2 = 3486.336

T2 = 31.2 °C

The final temperature of both substances when they reach thermal equilibrium is 31.2 °C

4 0

3 years ago

Carbonic anhydrase of erythrocytes (Mr 30,000) has one of the highest turnover numbers known. It catalyzes the reversible hydrat

Afina-wow [57]

Explanation:

According to the given data, the turnover number can be calculated as follows.

Turnover number = $K_{cat} = \frac{V_{max}}{\text{Concentration of enzyme}}$

$V_{max} = \frac{\text{Moles of CO_{2} hydrolyzed}{second}$

Therefore, moles of $CO_{2}$ hydrolyzed is as follows.

Moles of $CO_{2}$ hydrolyzed = $\frac{Mass of CO_{2}}{Molar mass of CO_{2}}$

= $\frac{0.30}{44}$

= 0.00682 moles

Now, moles of $CO_{2}$ hydolyzed per second is calculated as follows.

Moles of $CO_{2}$ hydolyzed per second = $\frac{0.00682}{60}$

= $1.137 \times 10^{-4} moles/second = V_{max}$

And,

Moles of enzyme = $\frac{Mass}{\text{Molar mass}}$

= $\frac{10.0 \mu g}{30000}$

= $3.33 \times 10^{-10}$ moles

Therefore, the value of $K_{cat}$ is as follows.

$K_{cat} = \frac{1.137 \times 10^{-4} moles}{3.333 \times 10^{-10} moles}$

= $0.3411 \times 10^{6}$ per second

= $0.3411 \times 60 \times 10^{6}$ per minute

= $20.466 \times 10^{6}$ per minute

Thus, we can conclude that the turnover number ( $K_{cat}$ ) of carbonic anhydrase (in units of $min^{-1}$ ) is $20.466 \times 10^{6}$ per minute.

6 0

4 years ago