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Tasya [4]
3 years ago
15

Sarah used 2.5 cups of cheese in a dish that serves 10 people. Arun used 1.6 cups of cheese in a dish that serves 8 people. How

much more cheese is in one serving of Sarah's dish? *
Chemistry
1 answer:
navik [9.2K]3 years ago
3 0

Answer:

0.05

Explanation:

We find the rate of the chess

Sarah is 2.5/10

Arun is 1.6/8

We will get

0.25 for Sarah

0.20 for Arun

We subtract the two

0.25 - 0.20, we then get 0.05

Sarah has 0.05 more cheese in one serving than Arun

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What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

6 0
4 years ago
Which statement is true about the concentration of the reactants and products in chemical equilibrium?
bulgar [2K]

Answer:

Ans: 2

Explanation:

The concentration of reactants and the concentration of products are constant.  

5 0
3 years ago
Sperm can be carried by _____ grains. ANSWERSSSSSSSSSSSSSSSSS O spore O stamen O pollen
7nadin3 [17]

What is pollen in the reproduction cycle of flowering plants?

A pollen grain is a microspore containing the male gametophyte, usually reduced to two undivided cells, each with one haploid (n) nucleus. These cells are surrounded by a very resistant wall, the exine, which generally has apertures, zones with less resistance which will allow the germination of the pollen tube.

Explanation of the reproduction cycle (cf diagram above)

A given species produces flowers bearing stamens. According to species, these flowers can be unisexual (stamens only) or bisexual (stamens and carpels).

The stamen anthers include 4 pollen sacs containing sporogenous cells (diploid=2n). These sporogenous cells undergo meiosis, each producing 4 microspores (haploid=n). Two nuclei are then formed by mitosis : the vegetative nucleus and the generative nucleus. The latter divides to form 2 sperms. Simultaneously the wall of the microspores becomes thicker and takes the characteristic shape of the species : it is a pollen grain (see: What are the morphological characteristics of pollen and spore grains?). In the majority of species, the 4 grains (resulting from the 4 microspores) split up into single grains; in some cases, they remain together (tetrad = group of 4 grains). When mature, pollen grains are released by the opening of the anthers.

A pollen grain is aimed at reaching another flower of the same species, bearing carpels. The ovaries contain ovules, in which meiosis occurs, then mitoses. It results in an embryo sac with 8 nuclei, among which an egg cell and 2 central cells. When a pollen grain arrives on another flower (see : How are the spores and pollen grains transported?), it is received by the stigmas.

The pollen grain germinates through an opening of the wall: the vegetative nucleus develops into a pollen tube which is guided by the style to the ovary, then enters the micropyle of an ovule. The pollen tube releases 2 sperm nuclei into the ovule: one of the sperms fuses with the egg cell into a zygote (2n), while the other sperm fuses with central nuclei and gives rise to albumen (= food source). There are generally several ovules in an ovary : each one can be fertilized by a distinct pollen grain.

Each fertilized ovule and its albumen form a seed that will develop into a new individual of this species. hope it works

4 0
3 years ago
Read 2 more answers
Consider separate solutions of NaOH and KClmade by dissolving 100.0 g of each solute in 250.0 mL of solution. Calculate the conc
gladu [14]

Answer:

Explanation:

MW of NaOH = 40 g/mol

MW of KCl = 74.55 g/mp;

250 mL = .25 L

100g NaOH / 40 g = 25 mol

100g KCl/ 74.55g = 1.34 mol

Molarity of NaOH: 25/.25 = 100M

Molarity of KCl: 1.34/.25 = 5.36 M

8 0
3 years ago
When 7.085 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 21.71 grams of CO2 and 10.37 grams of H
Taya2010 [7]

Answer:

- Empirical:

C_3H_7

- Molecular:

C_6H_{14}

Explanation:

Hello,

In this case, based on the information regarding the combustion, the moles of carbon turn out:

n_C=21.71gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}=0.493molC

Moreover, the moles of hydrogen:

n_H=10.37gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O}=1.152molH

Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:

C=\frac{0.4934}{0.4934}=1\\H=\frac{1.15222}{0.4934}=2.335\\CH_{2.335}

Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:

C_3H_7

In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:

C_6H_{14}

Which is hexane.

Best regards.

6 0
4 years ago
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