Milliy-sertifikat
HomeSubjectsMaterialsTestsRankingPricing
Sign inGet started
Main menu
Profile
GuestUse your Google account to sign in
SectionsHomeStudy planWritten partAssistantSubjectsMaterialsTestsRankingPricingHow it worksExam datesResultsFAQ
SettingsSettings
Milliy-sertifikatAn independent preparation platform. Official results and registration live on the government portals.
Telegram: @milliysertifikatInstagram
ExamWhat the National Certificate isScores and gradesExam dates
PlatformSubjectsMaterialsTestsRankingHow it worksPricing
CompanyResultsFAQContactTerms of usePrivacy policyMoney-back guarantee
© 2026 Milliy-sertifikat.uz. All rights reserved.
Written part

Written tasks and essays — checked by AI

Chemistry and biology tasks 41–43 and the native-language essay, by the official criteria. Write your solution — the AI scores every part and tells you what is missing.

Chemistry

24 tasks
  1. 01A closed 2 L vessel was charged with 0.6 mol of nitrogen and 1.8 mol of hydrogen and heated with a catalyst: N₂ + 3H₂ ⇌ 2NH₃. At equilibrium the total amount of gas had decreased by 20 % compared with the initial amount (constant temperature).25 points
  2. 02In the lab, 400 g of an 8 % copper(II) sulfate solution (density 1.084 g/mL) was prepared from blue vitriol (CuSO₄·5H₂O) and water. The solution was then electrolysed with inert electrodes at 5 A, and electrolysis was stopped when the mass of the solution had decreased by 12 g.25 points
  3. 03Complete combustion of 5.6 L (STP) of a methane–ethane mixture (liquid water formed) released 289.515 kJ of heat. Enthalpies of formation (kJ/mol): ΔfH°(CH₄) = −74.8; ΔfH°(C₂H₆) = −84.7; ΔfH°(CO₂) = −393.5; ΔfH°(H₂O, l) = −285.8.25 points
  4. 04The reaction 2NO(g) + O₂(g) → 2NO₂(g) is treated as elementary, i.e. its rate obeys the law of mass action. Initially [NO] = 0.4 mol/L and [O₂] = 0.3 mol/L; after 20 s the NO concentration fell to 0.2 mol/L. The temperature coefficient of the reaction is γ = 2.5.25 points
  5. 05Natural chlorine consists of the isotopes ³⁵Cl and ³⁷Cl; its mean relative atomic mass is 35.45 (take the isotopic masses equal to the mass numbers). A 22.18 g sample of calcium chloride made from natural chlorine is also given (Ar(Ca) = 40).25 points
  6. 06The dissociation constant of acetic acid is Ka = 1.8·10⁻⁵ (25 °C). Treat the weak electrolyte with α ≪ 1. Helpful values: √1.8 ≈ 1.342; lg 1.342 ≈ 0.128; lg 4.243 ≈ 0.628; lg 1.8 ≈ 0.255; lg 5.27 ≈ 0.722; Kw = 1·10⁻¹⁴.25 points
  7. 07When 7.8 g of a magnesium–aluminium mixture was completely dissolved in excess hydrochloric acid, 8.96 L (STP) of hydrogen was evolved.25 points
  8. 08A mixture of carbon(II) oxide and carbon(IV) oxide has a density relative to helium of 9. 11.2 L (STP) of the mixture is taken.25 points
  9. 09The following sequence is given:25 points
  10. 10Among the compounds with formula C₄H₈O₂ there are carboxylic acids and esters. Some give the silver-mirror test, others release a gas with sodium hydrogen carbonate.25 points
  11. 112-Methylbutane was reacted with bromine in light (1 : 1 mole ratio). The main product was isolated and heated with alcoholic potassium hydroxide.25 points
  12. 12The following sequence is given:25 points
  13. 13Glucose is the most abundant monosaccharide. It gives reactions typical both of aldehydes and of polyhydric alcohols, and in solution exists mainly in cyclic forms. Fermentation of glucose yields various substances.25 points
  14. 14Three isomers A, B and C of composition C₃H₈O are given. A and B release hydrogen with sodium; C does not react with sodium. Hot copper(II) oxide converts A into a substance that gives the silver-mirror test and B into one that does not. Heating A or B with concentrated sulfuric acid at 170 °C gives the same gaseous product.25 points
  15. 15Aniline is an important raw material for dyes and drugs. The transformations are:25 points
  16. 16Polymers are everywhere: polyethylene bags, PVC pipes, polystyrene containers, rubber tyres, capron (nylon-6) and lavsan (PET) fibres. They are made by addition polymerisation or polycondensation.25 points
  17. 1711.2 g of iron was dissolved in excess dilute sulfuric acid. A 5 % potassium permanganate solution was then added (in the presence of sulfuric acid) until it stopped being decolourised. Excess sodium hydroxide solution was then added; the precipitate was separated and heated to constant mass.25 points
  18. 18Two identical 19.2 g copper samples were taken. The first was completely dissolved in excess concentrated nitric acid, the second in excess dilute nitric acid. The salt from the second experiment was isolated and heated to constant mass, and the evolved gases were completely absorbed in 200 g of water.25 points
  19. 19Sodium hydroxide solution was added stepwise to a solution containing 26.7 g of aluminium chloride: first 0.6 mol, then another 0.1 mol of NaOH. Alkali was then added until the precipitate dissolved completely, and carbon dioxide was passed through the solution until no more precipitate formed. The precipitate was separated and heated.25 points
  20. 20Excess concentrated hydrochloric acid acted on 15.8 g of potassium permanganate. The chlorine evolved was split into two equal parts. The first was passed through hot (excess) potassium hydroxide solution; the oxygen-containing salt formed was isolated and heated with MnO₂. The second part reacted with excess heated iron.25 points
  21. 21To model sulfuric acid production, 24 g of pyrite (FeS₂) was completely roasted in excess air. The sulfur(IV) oxide formed was oxidised by oxygen over a catalyst (80 % yield), and the sulfur(VI) oxide obtained was completely absorbed in 100 g of 9.8 % sulfuric acid.25 points
  22. 22Chemical analysis. A solution brought to the laboratory is thought to contain barium, iron(III), copper(II) and ammonium nitrates. Adding excess dilute sulfuric acid to 100 mL of it gave 2.33 g of a white crystalline precipitate insoluble in acids. Excess ammonia solution was added to one portion of the filtrate; other portions were tested with separate reagents.25 points
  23. 23Chemical analysis (permanganatometry). A sample containing an iron(II) salt was dissolved in dilute sulfuric acid and made up to the mark in a 250 mL volumetric flask. A 20.00 mL aliquot was titrated, in the presence of sulfuric acid, with 0.0200 mol/L potassium permanganate; 16.00 mL of titrant was used to the equivalence point.25 points
  24. 24Chemical analysis. Four unlabelled tubes contain solutions of sodium chloride, sodium iodide, sodium carbonate and sodium sulfate. Only dilute nitric acid, barium chloride, silver nitrate and ammonia solution are available. The sodium chloride solution was made by dissolving 1.17 g of the salt.25 points

Native language and literature

16 tasks
  1. 01Write your response to the following situation.24 points
  2. 02Write your response to the following situation.24 points
  3. 03Write your response to the following situation.24 points
  4. 04Write your response to the following situation.24 points
  5. 05Write your response to the following situation.24 points
  6. 06Write your response to the following situation.24 points
  7. 07Write your response to the following situation.24 points
  8. 08Write your response to the following situation.24 points
  9. 09Write your response to the following situation.24 points
  10. 10Write your response to the following situation.24 points
  11. 11Write your response to the following situation.24 points
  12. 12Write your response to the following situation.24 points
  13. 13Write your response to the following situation.24 points
  14. 14Write your response to the following situation.24 points
  15. 15Write your response to the following situation.24 points
  16. 16Write your response to the following situation.24 points

Biology

24 tasks
  1. 01A double-stranded DNA fragment contains 1500 nucleotides, 20 % of which are adenine nucleotides. The fragment has no introns; one strand fully encodes a protein and its last triplet is a stop codon. The distance between adjacent base pairs is 0.34 nm, the mean nucleotide mass is 345 Da and an amino acid 110 Da (ignore water released on peptide-bond formation).30 points
  2. 02During intense running, a total of 10 mol of glucose was broken down in an athlete’s muscle cells and 236 mol of ATP was synthesised. Part of the glucose was broken down completely (through the oxygen stage) and the rest, for lack of oxygen, only by glycolysis to lactic acid. Data: 1 mol of glucose releases 200 kJ in glycolysis and 2800 kJ when broken down completely; 1 mol of ATP stores 40 kJ.30 points
  3. 03Cat somatic cells contain 38 chromosomes (2n = 38). Spermatogenesis is studied in a tomcat’s testis; 50 primary spermatocytes were selected.30 points
  4. 04Phenylketonuria is inherited as an autosomal recessive trait (a — disease allele). In a city of 500,000 people the disease occurs in one of every 10,000 newborns. Treat the population as ideal (random mating, large, no migration or selection).30 points
  5. 05A steppe ecosystem has the food chain: grasses → grasshoppers → frogs → snakes → eagle. In one season the eagle’s body mass increased by 6 kg (assume entirely through this chain). Grass productivity is 2 kg/m² per year; a frog weighs 50 g on average, a grasshopper 2 g.30 points
  6. 06In one day a wheat plant absorbed 264 g of carbon dioxide in photosynthesis and converted all the glucose formed into starch (gas volumes at STP).30 points
  7. 07An mRNA fragment used for polypeptide synthesis: 5′-AUG GCU UUC AAA GGC UGG UAA-3′.30 points
  8. 08An enzyme in a bacterial cell has a molecular mass of 44,000 Da. It consists of one polypeptide chain and is encoded by an intron-free gene (the coding part includes the stop codon). Mean amino-acid mass is 110 Da, nucleotide mass 345 Da, and the distance between adjacent nucleotides 0.34 nm (ignore water loss).30 points
  9. 09In peas, yellow seed colour (A) is completely dominant over green (a) and round seed shape (B) over wrinkled (b); the genes are on different chromosome pairs. A homozygous yellow round plant was crossed with a green wrinkled plant. F₁ plants were intercrossed, giving 1600 F₂ plants.35 points
  10. 10In humans, haemophilia is caused by a recessive X-linked gene (h) and albinism by a recessive autosomal gene (a). A woman with normal pigmentation and normal blood clotting had a haemophiliac father and an albino mother. She married a man with normal clotting and normal pigmentation whose mother is albino.35 points
  11. 11In Drosophila, grey body (B) is dominant over black (b) and normal wings (V) over vestigial (v); both genes are on one autosome. A grey normal-winged female heterozygous for both genes (dominant alleles from her father, recessive from her mother) was crossed with a black vestigial male. The offspring were 415 grey normal, 405 black vestigial, 92 grey vestigial and 88 black normal flies.35 points
  12. 12In humans the ABO blood groups are determined by three alleles of one gene (I⁰, Iᴬ, Iᴮ) and the Rh factor by an autosomal dominant gene (R — Rh-positive, r — Rh-negative); the genes are on different chromosomes. An Rh-positive woman of group II (A) had a father of group I (O) who was Rh-negative. She married an Rh-negative man of group III (B) whose mother was group I (O).35 points
  13. 13In four-o’clock (Mirabilis), flower colour shows incomplete dominance: RR — red, Rr — pink, rr — white. Tall stem (D) is completely dominant over short (d); the genes are on different chromosomes. A homozygous red tall plant was crossed with a white short plant. F₁ plants were intercrossed, giving 3200 F₂ plants.35 points
  14. 14In sweet pea, purple flowers appear only when two dominant genes (C and P) are both present; if either is missing, or both are recessive, flowers are white. The genes are on different chromosomes. Crossing two different homozygous white varieties (CCpp and ccPP) gave an all-purple F₁; intercrossing F₁ gave 1600 F₂ plants.35 points
  15. 15In cats, the genes for black (Xᴮ) and ginger (Xᵇ) coat colour are on the X chromosome and are codominant: XᴮXᵇ females are tortoiseshell. A tortoiseshell female was crossed with a ginger male.35 points
  16. 16Red grain colour in wheat is determined by two pairs of polymeric (cumulative) genes: each dominant allele (A₁, A₂) adds equally to colour intensity; a₁a₁a₂a₂ grain is white. The genes are on different chromosomes. A dark-red variety (A₁A₁A₂A₂) was crossed with a white variety; F₁ was intercrossed, giving 1600 F₂ plants.35 points
  17. 17In a cattle herd, polled (A) is completely dominant over horned (a). In a herd of 2500, horned animals make up 16 %; treat the herd as a randomly mating population.10 points
  18. 18A fragment of the coding strand of a DNA double helix: 5′-ATG CGT TAC GGA-3′. Adjacent nucleotides are 0.34 nm apart.10 points
  19. 19Under oxygen shortage, 5 mol of glucose in muscle cells was broken down only by glycolysis to lactic acid. 1 mol of glucose gives 200 kJ in glycolysis; 1 mol of ATP stores 40 kJ.10 points
  20. 20In humans, polydactyly is an autosomal dominant trait (gene P) with 80 % penetrance. Both parents are heterozygous (Pp).10 points
  21. 21A lake food chain: phytoplankton → zooplankton → small fish → pike. Without fishing, the total pike biomass can grow by 8 kg in a season. On average 0.4 kg of phytoplankton is produced per 1 m³ of water.10 points
  22. 22In a woman’s ovary, 20 primary oocytes completed meiosis (assume the first polar body also divided). In humans 2n = 46.10 points
  23. 23A bacterial cell synthesised a polypeptide of 150 amino acids. The gene encoding it has no introns and includes a stop codon; adjacent nucleotides are 0.34 nm apart.10 points
  24. 24In tomato, tall stem (A) is dominant over short (a) and round fruit (B) over pear-shaped (b); the genes are on one chromosome, 10 map units apart. An AB/ab plant was crossed with an ab/ab plant, giving 400 plants.10 points